OP Malhotra Class 10 Maths Solutions Chapter 12 Similar Triangles Exercise 12 (D)

S Chand Class 10 ICSE Maths Solutions Chapter 12 Similar Triangles Ex 12(d)

 

Question 1. In the figure, in ΔPQR if XY || QR, PX = 1 cm, XQ = 3 cm, YR = 4.5 cm and QR = 9 cm, find (i) PY and (ii) XY. Further, if the area of the ∆PXY is A cm², find in terms of A, (iii) the area of the triangle PQR (iv) the area of the figure XYRQ.

Answer:In triangle PQR, we are given that XY is parallel to QR. This means that triangle PXY is similar to triangle PQR (by AA similarity, since \( \angle P \) is common and \( \angle PXY = \angle PQR \) and \( \angle PYX = \angle PRQ \) as corresponding angles). Therefore, the ratio of their corresponding sides is equal: \( \frac{PX}{PQ} = \frac{PY}{PR} = \frac{XY}{QR} \) We are given: \( PX = 1 \) cm \( XQ = 3 \) cm \( YR = 4.5 \) cm \( QR = 9 \) cm First, let's find the full lengths of the sides of the larger triangle: \( PQ = PX + XQ = 1 + 3 = 4 \) cm \( PR = PY + YR \) Now we can use the ratio of sides. Let \( PY = x \) cm and \( XY = y \) cm. \( \frac{PX}{PQ} = \frac{PY}{PR} = \frac{XY}{QR} \) Substitute the known values: \( \frac{1}{4} = \frac{x}{x + 4.5} = \frac{y}{9} \) To find \( x \) (PY): \( \frac{1}{4} = \frac{x}{x + 4.5} \) Cross-multiply: \( 1 \times (x + 4.5) = 4 \times x \) \( x + 4.5 = 4x \) Subtract \( x \) from both sides: \( 4.5 = 4x - x \) \( 4.5 = 3x \) Divide by 3: \( x = \frac{4.5}{3} \) \( x = 1.5 \) cm So, (i) \( PY = 1.5 \) cm. To find \( y \) (XY): \( \frac{1}{4} = \frac{y}{9} \) Cross-multiply: \( 1 \times 9 = 4 \times y \) \( 9 = 4y \) Divide by 4: \( y = \frac{9}{4} \) \( y = 2.25 \) cm So, (ii) \( XY = 2.25 \) cm. Now for the areas. Since \( \triangle PXY \sim \triangle PQR \), the ratio of their areas is equal to the square of the ratio of their corresponding sides. \( \frac{\text{Area}(\triangle PXY)}{\text{Area}(\triangle PQR)} = \left( \frac{PX}{PQ} \right)^2 \) We know \( PX = 1 \) cm and \( PQ = 4 \) cm. \( \frac{\text{Area}(\triangle PXY)}{\text{Area}(\triangle PQR)} = \left( \frac{1}{4} \right)^2 = \frac{1}{16} \) We are given that \( \text{Area}(\triangle PXY) = A \) cm². \( \frac{A}{\text{Area}(\triangle PQR)} = \frac{1}{16} \) Cross-multiply: \( \text{Area}(\triangle PQR) = 16A \) cm² So, (iii) the area of the triangle PQR is \( 16A \) cm². Finally, for the area of the figure XYRQ, which is a trapezium. This area can be found by subtracting the area of the smaller triangle PXY from the area of the larger triangle PQR. \( \text{Area}(\text{XYRQ}) = \text{Area}(\triangle PQR) - \text{Area}(\triangle PXY) \) \( \text{Area}(\text{XYRQ}) = 16A - A \) \( \text{Area}(\text{XYRQ}) = 15A \) cm² So, (iv) the area of the figure XYRQ is \( 15A \) cm². Similar triangles have proportional areas, a useful property in geometry.In simple words: First, we find the missing lengths of the small triangle's sides by using the rule that similar triangles have sides that are in equal proportion. Then, we use the rule that the areas of similar triangles are related by the square of their side ratio to find the area of the big triangle. Finally, we subtract the small triangle's area from the big one to get the area of the four-sided shape at the bottom.

🎯 Exam Tip: Remember to clearly state the similarity criterion (AA, SSS, SAS) used for triangles and correctly identify corresponding sides for ratio calculations. For areas, the ratio is the square of the side ratio.

 

Question 2. The figure shows two isosceles similar triangles. If PQ and BC are not parallel and PC = 4cm, AQ = 3cm, QB = 12cm and BC = 15cm, calculate :
(i) the length of AP.
(ii) the ratio of the areas of triangle APQ

Answer:We are given that \( \triangle ABC \sim \triangle APQ \). Note that the problem statement for Question 2 mentions "two isosceles similar triangles", but the solution identifies \( \triangle ABC \sim \triangle APQ \). The diagram shows P on AC and Q on AB. Given values: \( PC = 4 \) cm \( AQ = 3 \) cm \( QB = 12 \) cm \( BC = 15 \) cm Let \( AP = x \) cm and \( PQ = y \) cm. From the similarity \( \triangle ABC \sim \triangle APQ \), the ratios of corresponding sides are equal. Based on the vertex correspondence in the similarity statement and the visual in the solution: \( \frac{AQ}{AC} = \frac{AP}{AB} = \frac{PQ}{BC} \) First, let's find the full lengths of sides AC and AB: \( AB = AQ + QB = 3 + 12 = 15 \) cm \( AC = AP + PC = x + 4 \) cm Now substitute these into the ratio equation: \( \frac{3}{x+4} = \frac{x}{15} = \frac{y}{15} \) (i) To find the length of AP (which is \( x \)): We use the first two parts of the ratio: \( \frac{3}{x+4} = \frac{x}{15} \) Cross-multiply: \( 3 \times 15 = x \times (x+4) \) \( 45 = x^2 + 4x \) Rearrange into a quadratic equation: \( x^2 + 4x - 45 = 0 \) Now, factor the quadratic equation. We need two numbers that multiply to -45 and add up to 4. These numbers are 9 and -5. \( x^2 + 9x - 5x - 45 = 0 \) Factor by grouping: \( x(x+9) - 5(x+9) = 0 \) \( (x+9)(x-5) = 0 \) This gives two possible values for \( x \): \( x+9 = 0 \implies x = -9 \) \( x-5 = 0 \implies x = 5 \) Since length cannot be negative, we take the positive value. So, \( AP = x = 5 \) cm. (ii) To find the ratio of the areas of triangle APQ to triangle ABC. Since \( \triangle APQ \sim \triangle ABC \), the ratio of their areas is equal to the square of the ratio of their corresponding sides. We can use the ratio \( \frac{AP}{AB} \) or \( \frac{AQ}{AC} \) or \( \frac{PQ}{BC} \). Using \( AP = 5 \) cm and \( AB = 15 \) cm: \( \frac{\text{Area}(\triangle APQ)}{\text{Area}(\triangle ABC)} = \left( \frac{AP}{AB} \right)^2 = \left( \frac{5}{15} \right)^2 \) Simplify the fraction inside the parenthesis: \( \left( \frac{5}{15} \right)^2 = \left( \frac{1}{3} \right)^2 = \frac{1}{9} \) The ratio between the areas of \( \triangle APQ \) and \( \triangle ABC \) is \( 1:9 \). This is a general property of similar figures where the ratio of areas is the square of the ratio of corresponding linear dimensions.In simple words: First, we use the fact that the two triangles are similar, which means their sides are proportional. We set up an equation to find the unknown length of side AP. Since length must be positive, we choose the correct value. Then, to find the ratio of their areas, we use the rule that the area ratio is the square of the side ratio.

🎯 Exam Tip: When solving for lengths using quadratic equations, always discard negative solutions as length cannot be negative. For similar triangles, remember the area ratio is the square of the side ratio, not just the ratio itself.

 

Question 3. DEF are similar. Find
(a) the ratio of the area of ∆ABC to the area of ADEF if AB = 2 and DE = 4.
(b) \( \frac { AB }{ DE } \), if ΔΑΒC : ADEF = 16 : 25.

Answer:Given that \( \triangle ABC \sim \triangle DEF \). (a) If \( AB = 2 \) and \( DE = 4 \): The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. \( \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left( \frac{AB}{DE} \right)^2 \) Substitute the given side lengths: \( \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left( \frac{2}{4} \right)^2 \) Simplify the fraction: \( \left( \frac{1}{2} \right)^2 = \frac{1}{4} \) So, the ratio between their areas is \( 1:4 \). This means the larger triangle's area is four times that of the smaller one because its sides are twice as long. (b) If \( \text{Area}(\triangle ABC) : \text{Area}(\triangle DEF) = 16 : 25 \): We use the same property: \( \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left( \frac{AB}{DE} \right)^2 \) Substitute the given area ratio: \( \frac{16}{25} = \left( \frac{AB}{DE} \right)^2 \) To find the ratio \( \frac{AB}{DE} \), take the square root of both sides: \( \sqrt{\frac{16}{25}} = \frac{AB}{DE} \) \( \frac{4}{5} = \frac{AB}{DE} \) So, \( \frac{AB}{DE} = \frac{4}{5} \). The side ratio is the square root of the area ratio.In simple words: When two triangles are similar, the ratio of their areas is found by squaring the ratio of their matching sides. If you know the side ratio, you square it to get the area ratio. If you know the area ratio, you take its square root to find the side ratio.

🎯 Exam Tip: Always remember that area ratios are related by the *square* of the side ratios. A common mistake is to simply use the side ratio directly for areas.

 

Question 4. In a ∆ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point on AC such that PQ || BC. Find the ratio of the areas of ∆APQ and trapezium BPQC.

Answer:In \( \triangle ABC \), we are given that P is a point on AB and Q is a point on AC such that PQ is parallel to BC. We are also given the ratio \( AP : PB = 1 : 2 \). From this ratio, we can find the ratio of AP to the whole side AB: \( AP = 1k \), \( PB = 2k \) \( AB = AP + PB = 1k + 2k = 3k \) So, \( \frac{AP}{AB} = \frac{1k}{3k} = \frac{1}{3} \) Since PQ || BC, by the Basic Proportionality Theorem or by AA similarity (as \( \angle A \) is common and corresponding angles are equal), \( \triangle APQ \sim \triangle ABC \). The ratio of their areas is the square of the ratio of their corresponding sides: \( \frac{\text{Area}(\triangle APQ)}{\text{Area}(\triangle ABC)} = \left( \frac{AP}{AB} \right)^2 \) Substitute the ratio \( \frac{AP}{AB} = \frac{1}{3} \): \( \frac{\text{Area}(\triangle APQ)}{\text{Area}(\triangle ABC)} = \left( \frac{1}{3} \right)^2 = \frac{1}{9} \) Let's assume \( \text{Area}(\triangle APQ) = 1 \) unit. Then \( \text{Area}(\triangle ABC) = 9 \) units. The figure BPQC is a trapezium formed by removing \( \triangle APQ \) from \( \triangle ABC \). So, \( \text{Area}(\text{trapezium BPQC}) = \text{Area}(\triangle ABC) - \text{Area}(\triangle APQ) \) \( \text{Area}(\text{trapezium BPQC}) = 9 - 1 = 8 \) units. Now, we need to find the ratio of the areas of \( \triangle APQ \) and trapezium BPQC: \( \text{Area}(\triangle APQ) : \text{Area}(\text{trapezium BPQC}) = 1 : 8 \) So, the ratio is \( 1:8 \). Understanding how areas combine or subtract is key here.In simple words: We first figure out the ratio of the smaller triangle's side (AP) to the larger triangle's side (AB). Because the line PQ is parallel to BC, the small triangle APQ is similar to the big triangle ABC. This means their areas are related by the square of their side ratio. Once we know the small triangle's area in relation to the big one, we subtract to find the area of the bottom part, which is the trapezium, and then we state the required ratio.

🎯 Exam Tip: When a line is drawn parallel to one side of a triangle, cutting the other two sides, it creates a smaller triangle similar to the original. This is a very common setup for questions involving ratios of areas and sides.

 

Question 5. The areas of two similar triangles ABC and DEF are 36 cm² and 81 cm² respectively. If EF = 6.9 cm, determine BC.
Answer:We are given that \( \triangle ABC \sim \triangle DEF \). The areas are: \( \text{Area}(\triangle ABC) = 36 \) cm² \( \text{Area}(\triangle DEF) = 81 \) cm² The length of one side of \( \triangle DEF \) is \( EF = 6.9 \) cm. We need to find the corresponding side BC in \( \triangle ABC \). For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides. \( \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left( \frac{BC}{EF} \right)^2 \) Substitute the given values: \( \frac{36}{81} = \left( \frac{BC}{6.9} \right)^2 \) To solve for BC, first take the square root of both sides: \( \sqrt{\frac{36}{81}} = \frac{BC}{6.9} \) \( \frac{6}{9} = \frac{BC}{6.9} \) Simplify the fraction \( \frac{6}{9} \) to \( \frac{2}{3} \): \( \frac{2}{3} = \frac{BC}{6.9} \) Now, multiply both sides by 6.9 to find BC: \( BC = \frac{2}{3} \times 6.9 \) \( BC = 2 \times \frac{6.9}{3} \) \( BC = 2 \times 2.3 \) \( BC = 4.6 \) cm. The length of BC is \( 4.6 \) cm, demonstrating how side lengths scale with area.In simple words: Since the triangles are similar, we know that the ratio of their areas is the same as the square of the ratio of their matching sides. We set up this relationship using the given areas and the known side length. Then, we take the square root to find the ratio of the sides and solve for the unknown side BC.

🎯 Exam Tip: Ensure you correctly identify corresponding sides when setting up the ratio of areas. For example, if \( \triangle ABC \sim \triangle DEF \), then BC corresponds to EF, AB to DE, and AC to DF.

 

Question 6. In the figure, DE || BC. AD = 3 cm, DB = 2 cm, area of ∆ABC = 10 cm². Find the area of AADE.

Answer:In the given figure, we have triangle ADE with a line segment BC parallel to DE. We are given: \( DE \parallel BC \) (This implies that \( \triangle ABC \sim \triangle ADE \) if A, B, D are collinear and A, C, E are collinear, with ABC being the smaller triangle). \( AD = 3 \) cm \( DB = 2 \) cm \( \text{Area}(\triangle ABC) = 10 \) cm² First, we need to find the length of the corresponding side AB. From the given lengths \( AD = 3 \) cm and \( DB = 2 \) cm, and assuming that B lies between A and D (as implied by the solution's calculation of AB), we have: \( AB = AD - DB = 3 - 2 = 1 \) cm. This means the triangle ABC is smaller and nested within ADE, with A as the common vertex. Since \( BC \parallel DE \), we have \( \triangle ABC \sim \triangle ADE \) (by AA similarity, as \( \angle A \) is common and corresponding angles are equal). The ratio of the areas of similar triangles is equal to the square of the ratio of their corresponding sides: \( \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle ADE)} = \left( \frac{AB}{AD} \right)^2 \) Substitute the known values: \( \frac{10}{\text{Area}(\triangle ADE)} = \left( \frac{1}{3} \right)^2 \) \( \frac{10}{\text{Area}(\triangle ADE)} = \frac{1}{9} \) Cross-multiply to find \( \text{Area}(\triangle ADE) \): \( \text{Area}(\triangle ADE) = 10 \times 9 \) \( \text{Area}(\triangle ADE) = 90 \) cm². The area of \( \triangle ADE \) is \( 90 \) cm². This shows how significantly area can increase when side lengths are multiplied.In simple words: Because the line BC is parallel to DE, the smaller triangle ABC is similar to the larger triangle ADE. We first find the length of side AB. Then, we use the rule that the ratio of the areas of similar triangles is the square of the ratio of their matching sides. We use this to calculate the unknown area of triangle ADE.

🎯 Exam Tip: Carefully observe the points' arrangement (e.g., A-B-D or A-D-B) to correctly determine the total length of a side, especially when segments are given. This affects the side ratio. If parallel lines exist, AA similarity is often applicable.

 

Question 7. In the figure, the angles PRS and PQR are similar PS = 2 cm and PR = 3 cm. If the area of the triangle PRS is 2 cm², calculate the area of APQR.

Answer:We are given that \( \triangle PRS \) and \( \triangle PQR \) are similar triangles. The phrasing "the angles PRS and PQR are similar" likely means the triangles themselves are similar. Based on the solution's steps: We have \( \angle RPS \) (angle at P in \( \triangle PRS \)) which is common to \( \angle RPQ \) (angle at P in \( \triangle PQR \)). And we are given \( \angle PRS \sim \angle PQR \). Therefore, by the Angle-Angle (AA) axiom of similarity, \( \triangle PRS \sim \triangle PQR \). The vertex correspondence is: \( P \leftrightarrow P \), \( R (\text{of } \triangle PRS) \leftrightarrow Q (\text{of } \triangle PQR) \), \( S (\text{of } \triangle PRS) \leftrightarrow R (\text{of } \triangle PQR) \). The ratio of corresponding sides is: \( \frac{PS}{PR} = \frac{PR}{PQ} = \frac{RS}{QR} \) We are given: \( PS = 2 \) cm \( PR = 3 \) cm (This is the side PR of \( \triangle PRS \), and it corresponds to PQ of \( \triangle PQR \)). \( \text{Area}(\triangle PRS) = 2 \) cm² We can use the ratio of areas for similar triangles: \( \frac{\text{Area}(\triangle PRS)}{\text{Area}(\triangle PQR)} = \left( \frac{\text{corresponding side of } \triangle PRS}{\text{corresponding side of } \triangle PQR} \right)^2 \) Using the side correspondence \( PS \leftrightarrow PR \): \( \frac{\text{Area}(\triangle PRS)}{\text{Area}(\triangle PQR)} = \left( \frac{PS}{PR} \right)^2 \) Substitute the given values for sides and area: \( \frac{2}{\text{Area}(\triangle PQR)} = \left( \frac{2}{3} \right)^2 \) \( \frac{2}{\text{Area}(\triangle PQR)} = \frac{4}{9} \) Cross-multiply: \( 2 \times 9 = 4 \times \text{Area}(\triangle PQR) \) \( 18 = 4 \times \text{Area}(\triangle PQR) \) Divide by 4: \( \text{Area}(\triangle PQR) = \frac{18}{4} \) \( \text{Area}(\triangle PQR) = \frac{9}{2} \) \( \text{Area}(\triangle PQR) = 4.5 \) cm². The area of triangle PQR is \( 4.5 \) cm². Recognizing the correct vertex correspondence is critical for setting up the side ratios correctly.In simple words: We are told that two triangles, PRS and PQR, are similar. This means their angles are the same and their sides are in proportion. We use the rule that the ratio of their areas is the square of the ratio of their matching sides. Using the given side lengths and the area of the smaller triangle, we can calculate the area of the larger triangle.

🎯 Exam Tip: When two triangles share a common angle and have another pair of equal angles, they are similar by AA criterion. Always be careful to match the corresponding sides correctly based on the vertex order in the similarity statement.

 

Question 8. In the figure, DE BC and AD : DB = 5 : 4. Find \( \frac { \text{Area}(\triangle ADE) }{ \text{Area}(\triangle ABC) } \)

Answer:In \( \triangle ABC \), we are given that D is a point on AB and E is a point on AC. We are given that \( DE \parallel BC \). This implies that \( \triangle ADE \sim \triangle ABC \) (by AA similarity, as \( \angle A \) is common and corresponding angles are equal). We are given the ratio \( AD : DB = 5 : 4 \). This means \( AD = 5k \) and \( DB = 4k \) for some value \( k \). To find the total length of side AB: \( AB = AD + DB = 5k + 4k = 9k \) Now we can find the ratio of corresponding sides \( \frac{AD}{AB} \): \( \frac{AD}{AB} = \frac{5k}{9k} = \frac{5}{9} \) Since the triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides: \( \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = \left( \frac{AD}{AB} \right)^2 \) Substitute the ratio \( \frac{AD}{AB} \): \( \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = \left( \frac{5}{9} \right)^2 \) \( \frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle ABC)} = \frac{25}{81} \) The ratio of the areas of \( \triangle ADE \) to \( \triangle ABC \) is \( \frac{25}{81} \). This calculation is a direct application of the area similarity theorem.In simple words: Since the line DE is parallel to BC, the small triangle ADE is similar to the big triangle ABC. We first use the given ratio of AD to DB to find the ratio of AD to the whole side AB. Then, because the triangles are similar, the ratio of their areas is the square of the ratio of their matching sides, which allows us to find the desired area ratio.

🎯 Exam Tip: Always remember that when a line is parallel to one side of a triangle, it creates a smaller similar triangle. The ratio of areas follows the square of the ratio of corresponding sides.

 

Question 9. In the figure, in APQR, PQ = 8 cm, PR = 10 cm and ∠Q = 90°. A and B are points on sides PQ and PR respectively such that AB = 2 cm and ∠ABP = 90°. Find :
(i) the area of APAB
(ii) the area of quad. AQRB : area of APQR.

Answer:In \( \triangle PQR \), we are given: \( PQ = 8 \) cm \( PR = 10 \) cm \( \angle Q = 90^\circ \) (This means \( \triangle PQR \) is a right-angled triangle). Points A and B are on sides PQ and PR respectively. \( AB = 2 \) cm \( \angle ABP = 90^\circ \) Since \( \angle Q = 90^\circ \) and \( \angle ABP = 90^\circ \), this means AB is parallel to QR (as corresponding angles are equal if PQR is cut by line AB). Therefore, \( \triangle PAB \sim \triangle PQR \) (by AA similarity, as \( \angle P \) is common and \( \angle PAB = \angle PQR = 90^\circ \)). First, let's find the length of QR using the Pythagorean theorem in \( \triangle PQR \): \( PR^2 = PQ^2 + QR^2 \) \( 10^2 = 8^2 + QR^2 \) \( 100 = 64 + QR^2 \) \( QR^2 = 100 - 64 \) \( QR^2 = 36 \) \( QR = \sqrt{36} = 6 \) cm. (i) To find the area of \( \triangle PAB \): Since \( \triangle PAB \sim \triangle PQR \), the ratio of their corresponding sides is equal: \( \frac{AB}{QR} = \frac{PA}{PQ} = \frac{PB}{PR} \) We know \( AB = 2 \) cm and \( QR = 6 \) cm. So, \( \frac{AB}{QR} = \frac{2}{6} = \frac{1}{3} \) This means the ratio of similarity is \( \frac{1}{3} \). Now, using the side ratio \( \frac{PA}{PQ} = \frac{1}{3} \): \( \frac{PA}{8} = \frac{1}{3} \) \( PA = \frac{8}{3} \) cm. The area of a right-angled triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \). \( \triangle PAB \) is a right-angled triangle at A because \( \angle ABP = 90^\circ \) implies that if B is on PR and A on PQ, then PA is perpendicular to AB. However, the problem states \( \angle Q = 90^\circ \) and \( \angle ABP = 90^\circ \). If A is on PQ and B is on PR, and \( \angle ABP = 90^\circ \), this angle is at B, not A. Let's re-evaluate the right angle location for \( \triangle PAB \). If \( \triangle PAB \sim \triangle PQR \) and \( \angle Q = 90^\circ \), then the corresponding angle in \( \triangle PAB \) is \( \angle PAB \). So \( \angle PAB = 90^\circ \). Therefore, \( \triangle PAB \) is right-angled at A. Its area is \( \frac{1}{2} \times PA \times AB \). \( \text{Area}(\triangle PAB) = \frac{1}{2} \times \frac{8}{3} \times 2 \) \( \text{Area}(\triangle PAB) = \frac{8}{3} \) cm². The area of triangle PAB is \( \frac{8}{3} \) cm². (ii) To find the ratio of the areas of quadrilateral AQRB to \( \triangle PQR \): First, let's find the area of \( \triangle PQR \). Since it's right-angled at Q: \( \text{Area}(\triangle PQR) = \frac{1}{2} \times PQ \times QR \) \( \text{Area}(\triangle PQR) = \frac{1}{2} \times 8 \times 6 \) \( \text{Area}(\triangle PQR) = 24 \) cm². The quadrilateral AQRB is formed by subtracting \( \triangle PAB \) from \( \triangle PQR \). \( \text{Area}(\text{quad. AQRB}) = \text{Area}(\triangle PQR) - \text{Area}(\triangle PAB) \) \( \text{Area}(\text{quad. AQRB}) = 24 - \frac{8}{3} \) To subtract, find a common denominator: \( \text{Area}(\text{quad. AQRB}) = \frac{72}{3} - \frac{8}{3} = \frac{64}{3} \) cm². Now, find the ratio \( \frac{\text{Area}(\text{quad. AQRB})}{\text{Area}(\triangle PQR)} \): \( \frac{\text{Area}(\text{quad. AQRB})}{\text{Area}(\triangle PQR)} = \frac{64/3}{24} \) \( = \frac{64}{3 \times 24} = \frac{64}{72} \) Simplify the fraction by dividing both by 8: \( = \frac{8}{9} \) Alternatively, using the area ratio of similar triangles: \( \frac{\text{Area}(\triangle PAB)}{\text{Area}(\triangle PQR)} = \left( \frac{AB}{QR} \right)^2 = \left( \frac{2}{6} \right)^2 = \left( \frac{1}{3} \right)^2 = \frac{1}{9} \) So, \( \text{Area}(\triangle PAB) = \frac{1}{9} \times \text{Area}(\triangle PQR) \). If we let \( \text{Area}(\triangle PQR) = 9 \) units, then \( \text{Area}(\triangle PAB) = 1 \) unit. Then \( \text{Area}(\text{quad. AQRB}) = \text{Area}(\triangle PQR) - \text{Area}(\triangle PAB) = 9 - 1 = 8 \) units. The ratio \( \frac{\text{Area}(\text{quad. AQRB})}{\text{Area}(\triangle PQR)} = \frac{8}{9} \). Both methods yield the same result, confirming the calculation.In simple words: First, we use the Pythagorean theorem to find the missing side of the big right-angled triangle. Since the smaller triangle PAB has a right angle at B and shares an angle with the big triangle PQR, they are similar. We use their side ratio to find the area of the small triangle PAB. Then, we find the area of the big triangle PQR. Finally, we subtract the area of the small triangle from the big one to get the area of the four-sided shape (quadrilateral AQRB) and form the required ratio.

🎯 Exam Tip: When given right angles and points on sides, always check for parallel lines and similar triangles. Pythagorean theorem is often needed to find missing side lengths in right triangles. The area ratio of similar triangles can simplify calculations significantly.

 

Question 10. The areas of two similar triangles ABC and PQR are 64 sq. cm and 121 sq. cm respectively. If QR = 15.4 cm, find BC.
Answer:We are given that \( \triangle ABC \sim \triangle PQR \). The areas are: \( \text{Area}(\triangle ABC) = 64 \) cm² \( \text{Area}(\triangle PQR) = 121 \) cm² The length of one side of \( \triangle PQR \) is \( QR = 15.4 \) cm. We need to find the corresponding side BC in \( \triangle ABC \). For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides. \( \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \left( \frac{BC}{QR} \right)^2 \) Substitute the given values: \( \frac{64}{121} = \left( \frac{BC}{15.4} \right)^2 \) To solve for BC, first take the square root of both sides: \( \sqrt{\frac{64}{121}} = \frac{BC}{15.4} \) \( \frac{8}{11} = \frac{BC}{15.4} \) Now, multiply both sides by 15.4 to find BC: \( BC = \frac{8}{11} \times 15.4 \) \( BC = 8 \times \frac{15.4}{11} \) \( BC = 8 \times 1.4 \) \( BC = 11.2 \) cm. The length of BC is \( 11.2 \) cm. This calculation reinforces the relationship between areas and side lengths in similar figures.In simple words: We know that for similar triangles, the ratio of their areas is the square of the ratio of their matching sides. We use the given areas and the known side length to set up an equation. Then we take the square root to find the side ratio and solve for the unknown side BC.

🎯 Exam Tip: Always verify that you are taking the square root of the correct ratio. The larger area should correspond to the larger side, and vice-versa, when taking square roots.

 

Question 11. D, E, F are the mid-points of the sides BC, CA and AB respectively of a triangle ABC. Determine the ratio of the areas of the triangles DEF and AВС.

Answer:In \( \triangle ABC \), D, E, and F are the mid-points of sides BC, CA, and AB respectively. When the mid-points of the sides of a triangle are joined, they form four smaller congruent triangles, and the inner triangle (formed by joining midpoints) is similar to the original triangle. According to the Mid-point Theorem: 1. The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it. * Since E and F are mid-points of AC and AB, \( EF \parallel BC \) and \( EF = \frac{1}{2} BC \). So, \( \frac{EF}{BC} = \frac{1}{2} \) ... (i) * Since D and E are mid-points of BC and AC, \( DE \parallel AB \) and \( DE = \frac{1}{2} AB \). So, \( \frac{DE}{AB} = \frac{1}{2} \) ... (ii) * Since D and F are mid-points of BC and AB, \( DF \parallel AC \) and \( DF = \frac{1}{2} AC \). So, \( \frac{DF}{AC} = \frac{1}{2} \) ... (iii) From (i), (ii), and (iii), we see that the ratio of corresponding sides of \( \triangle DEF \) and \( \triangle ABC \) is \( \frac{1}{2} \): \( \frac{EF}{BC} = \frac{DE}{AB} = \frac{DF}{AC} = \frac{1}{2} \) Since all corresponding sides are in the same ratio, \( \triangle DEF \sim \triangle ABC \) by the SSS similarity criterion. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides: \( \frac{\text{Area}(\triangle DEF)}{\text{Area}(\triangle ABC)} = \left( \frac{EF}{BC} \right)^2 \) Substitute the side ratio \( \frac{EF}{BC} = \frac{1}{2} \): \( \frac{\text{Area}(\triangle DEF)}{\text{Area}(\triangle ABC)} = \left( \frac{1}{2} \right)^2 = \frac{1}{4} \) So, the ratio of the areas of \( \triangle DEF \) to \( \triangle ABC \) is \( 1:4 \). This property is a fundamental result of the midpoint theorem in geometry.In simple words: When you connect the mid-points of all three sides of a triangle, you create a smaller triangle inside. The Mid-point Theorem tells us that each side of this smaller triangle is half the length of the corresponding side of the original triangle. Because all sides are half, the small triangle is similar to the big one. The ratio of their areas is found by squaring this side ratio.

🎯 Exam Tip: The Mid-point Theorem is crucial for these types of questions. Remember that the triangle formed by joining midpoints is always similar to the original triangle with a side ratio of 1:2, and therefore an area ratio of 1:4.

 

Question 12. ABCD is a trapezium in which AB = 2 DC and AB || DC. If AC and BD intersect at O, prove that area (AAQB) = 4 area (ACOD).

Answer:Given that ABCD is a trapezium with \( AB \parallel DC \). We are also given \( AB = 2 DC \). The diagonals AC and BD intersect at point O. Consider \( \triangle AOB \) and \( \triangle COD \): 1. \( \angle AOB = \angle COD \) (Vertically opposite angles are equal). 2. Since \( AB \parallel DC \), and AC is a transversal, \( \angle OAB = \angle OCD \) (Alternate interior angles are equal). 3. Similarly, since \( AB \parallel DC \), and BD is a transversal, \( \angle OBA = \angle ODC \) (Alternate interior angles are equal). Therefore, \( \triangle AOB \sim \triangle COD \) by the Angle-Angle-Angle (AAA) similarity criterion. For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides: \( \frac{\text{Area}(\triangle AOB)}{\text{Area}(\triangle COD)} = \left( \frac{AB}{DC} \right)^2 \) We are given \( AB = 2 DC \). So, \( \frac{AB}{DC} = 2 \). Substitute this ratio into the area equation: \( \frac{\text{Area}(\triangle AOB)}{\text{Area}(\triangle COD)} = (2)^2 \) \( \frac{\text{Area}(\triangle AOB)}{\text{Area}(\triangle COD)} = 4 \) This means \( \text{Area}(\triangle AOB) = 4 \times \text{Area}(\triangle COD) \). Thus, we have proved that \( \text{Area}(\triangle AOB) = 4 \text{ Area}(\triangle COD) \). In the context of the question, AOB corresponds to the triangle AAQB, if Q is implied to be O. This property of trapeziums is a common geometry result.In simple words: In a trapezium where two sides are parallel, the triangles formed by the diagonals at their intersection point are similar. We first show that triangle AOB is similar to triangle COD using angle properties. Then, we use the given relationship between their parallel sides (AB is twice DC) and the rule that the area ratio of similar triangles is the square of their side ratio. This proves the required area relationship.

🎯 Exam Tip: When dealing with trapeziums and intersecting diagonals, always look for similar triangles. The triangles formed by the parallel sides and the point of intersection of diagonals are similar (e.g., \( \triangle AOB \sim \triangle COD \)). Alternate interior angles are key here.

 

Question 13. If the areas of two similar triangles are equal, prove that they are congruent.
Answer:Let the two similar triangles be \( \triangle ABC \) and \( \triangle DEF \). Given: \( \triangle ABC \sim \triangle DEF \). Also given: \( \text{Area}(\triangle ABC) = \text{Area}(\triangle DEF) \). Since the triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides. \( \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = \left( \frac{AB}{DE} \right)^2 = \left( \frac{BC}{EF} \right)^2 = \left( \frac{AC}{DF} \right)^2 \) Because \( \text{Area}(\triangle ABC) = \text{Area}(\triangle DEF) \), their ratio is 1: \( \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)} = 1 \) Substitute this into the area ratio equation: \( 1 = \left( \frac{AB}{DE} \right)^2 \implies \frac{AB}{DE} = \sqrt{1} = 1 \implies AB = DE \) \( 1 = \left( \frac{BC}{EF} \right)^2 \implies \frac{BC}{EF} = \sqrt{1} = 1 \implies BC = EF \) \( 1 = \left( \frac{AC}{DF} \right)^2 \implies \frac{AC}{DF} = \sqrt{1} = 1 \implies AC = DF \) So, we have found that \( AB = DE \), \( BC = EF \), and \( AC = DF \). This means all three corresponding sides of \( \triangle ABC \) and \( \triangle DEF \) are equal. By the Side-Side-Side (SSS) axiom of congruency, if all three sides of one triangle are equal to the corresponding three sides of another triangle, then the triangles are congruent. Therefore, \( \triangle ABC \cong \triangle DEF \). This proves that similar triangles with equal areas are identical.In simple words: We start with two triangles that are similar and have the same area. We know that the ratio of their areas is equal to the square of the ratio of their matching sides. Since their areas are equal, this ratio is 1. Taking the square root, we find that the ratio of their sides is also 1, meaning their matching sides are exactly equal. If all three sides of one triangle are equal to the three sides of another, then the triangles are congruent, meaning they are identical in size and shape.

🎯 Exam Tip: This is a standard theorem. Remember that similarity implies proportional sides, while congruency implies equal sides. The link between them when areas are equal is that the proportionality factor becomes 1.

 

Question 14. Prove that the ratio of corresponding altitudes of two similar triangles is equal to the ratio of their corresponding sides.

Answer:Let \( \triangle ABC \) and \( \triangle DEF \) be two similar triangles. Given: \( \triangle ABC \sim \triangle DEF \). This means their corresponding angles are equal, and the ratio of their corresponding sides is equal: \( \angle A = \angle D \), \( \angle B = \angle E \), \( \angle C = \angle F \) And \( \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} \) ... (1) Let AP be the altitude from A to BC in \( \triangle ABC \), where P is on BC. So \( AP \perp BC \). Let DQ be the altitude from D to EF in \( \triangle DEF \), where Q is on EF. So \( DQ \perp EF \). Now consider \( \triangle ABP \) and \( \triangle DEQ \): 1. \( \angle B = \angle E \) (Since \( \triangle ABC \sim \triangle DEF \), corresponding angles are equal). 2. \( \angle APB = \angle DQE = 90^\circ \) (Because AP and DQ are altitudes). Therefore, \( \triangle ABP \sim \triangle DEQ \) by the Angle-Angle (AA) similarity criterion. Since \( \triangle ABP \sim \triangle DEQ \), the ratio of their corresponding sides is equal: \( \frac{AB}{DE} = \frac{BP}{EQ} = \frac{AP}{DQ} \) ... (2) From equation (1), we know that the ratio of corresponding sides of the similar triangles \( \triangle ABC \) and \( \triangle DEF \) is \( \frac{AB}{DE} = \frac{BC}{EF} \). From equation (2), we found that \( \frac{AP}{DQ} = \frac{AB}{DE} \). Combining these results, we get: \( \frac{AP}{DQ} = \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} \) This means the ratio of the corresponding altitudes (AP and DQ) is equal to the ratio of their corresponding sides. This proves the theorem.In simple words: We start with two triangles that are similar. We draw altitudes (heights) in both triangles. This creates smaller right-angled triangles within each original triangle. We show that these smaller right-angled triangles are also similar because they share an angle and both have a right angle. Since they are similar, their sides are proportional. This means the ratio of their altitudes is the same as the ratio of their corresponding main sides, proving the rule.

🎯 Exam Tip: To prove relationships involving altitudes in similar triangles, create smaller right-angled triangles using the altitudes. Then, apply AA similarity to these smaller triangles to establish the proportionality of altitudes and sides.

 

Question 15. ABC is a triangle, PQ is a line segment intersecting AB in P and AC in Q such that PQ || BC and divides ΔΑΒC into two of equal parts equal area. Find \( \frac { BP }{ AB } \).

Answer:In \( \triangle ABC \), PQ is a line segment such that P is on AB and Q is on AC. Given: \( PQ \parallel BC \). This implies that \( \triangle APQ \sim \triangle ABC \) (by AA similarity, as \( \angle A \) is common and corresponding angles are equal). Also given: PQ divides \( \triangle ABC \) into two parts of equal area. This means \( \text{Area}(\triangle APQ) = \text{Area}(\text{quadrilateral PBCQ}) \). If these two parts are equal, then each part is half of the total area of \( \triangle ABC \): \( \text{Area}(\triangle APQ) = \frac{1}{2} \text{Area}(\triangle ABC) \) Now, we use the property that for similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides: \( \frac{\text{Area}(\triangle APQ)}{\text{Area}(\triangle ABC)} = \left( \frac{AP}{AB} \right)^2 \) Substitute \( \frac{\text{Area}(\triangle APQ)}{\text{Area}(\triangle ABC)} = \frac{1}{2} \): \( \frac{1}{2} = \left( \frac{AP}{AB} \right)^2 \) Take the square root of both sides: \( \frac{AP}{AB} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \) We need to find \( \frac{BP}{AB} \). We know that \( BP = AB - AP \). So, \( \frac{BP}{AB} = \frac{AB - AP}{AB} \) We can split this fraction: \( \frac{BP}{AB} = \frac{AB}{AB} - \frac{AP}{AB} \) \( \frac{BP}{AB} = 1 - \frac{AP}{AB} \) Substitute the value of \( \frac{AP}{AB} \): \( \frac{BP}{AB} = 1 - \frac{1}{\sqrt{2}} \) To rationalize and combine: \( \frac{BP}{AB} = \frac{\sqrt{2}}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} \) This is the required ratio \( \frac{BP}{AB} \). This calculation shows how areas and segment ratios are intertwined.In simple words: The line PQ is parallel to BC, so the small triangle APQ is similar to the big triangle ABC. We are told that PQ cuts the big triangle into two parts of equal area, meaning the small triangle's area is half of the big triangle's area. Using the rule that the area ratio of similar triangles is the square of their side ratio, we find the ratio of AP to AB. Finally, we use the fact that BP is the remaining part of AB after AP to find the ratio of BP to AB.

🎯 Exam Tip: Remember that if a line divides a triangle into two parts of equal area, the smaller triangle formed will have half the area of the original. When working with ratios involving parts of a side (like BP and AB), it's often easiest to find the ratio of the complete segment (AP to AB) first, then use it to find the ratio of the partial segment (BP to AB).

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