CBSE Class 10 Mathematics Triangles Assignment Set 13

Type I: Very Short Answer Type Questions (1 Mark Each)

Question 1. The lengths of the diagonals of a rhombus are 16 cm and 12 cm. Find the length of the side of the rhombus.
Answer:
The diagonals of a rhombus bisect each other at right angles, so each side is the hypotenuse of a right triangle with legs 8 cm and 6 cm.
Side \( = \sqrt{8^2 + 6^2} = \sqrt{100} = \) 10 cm.
In simple words: Half the diagonals are the legs of a right triangle.

Teacher's Note:
1. The diagonals of a rhombus meet at 90°.
2. Half-diagonals 8 and 6 form a 3-4-5 type triangle.

Teacher's Note:
1. The altitude to the hypotenuse splits the right triangle into two triangles similar to it.
2. Match the equal angles first, then write the sides.

Question 3. If in two triangles DEF and PQR, \( \angle D = \angle Q \) and \( \angle R = \angle E \), then show that \( ED \cdot PR = FE \cdot RQ \).
Answer:
Two angles of \( \triangle DEF \) equal two angles of \( \triangle QRP \): \( \angle D = \angle Q \) and \( \angle E = \angle R \). So \( \triangle DEF \sim \triangle QRP \) (AA).
Corresponding sides are proportional: \( \dfrac{DE}{QR} = \dfrac{EF}{RP} \), so \( ED \cdot PR = FE \cdot RQ \).
In simple words: The triangles match as D to Q, E to R, F to P.

Teacher's Note:
1. Write the similarity with the vertices in matching order.
2. Cross-multiply the proportional sides.

Question 4. In triangles ABC and DEF, \( \angle B = \angle E \), \( \angle F = \angle C \) and \( AB = 3DE \). Show that the two triangles are similar but not congruent.
Answer:
\( \angle B = \angle E \) and \( \angle C = \angle F \), so \( \triangle ABC \sim \triangle DEF \) (AA).
Then \( \dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF} = 3 \), so every side of \( \triangle ABC \) is 3 times the matching side of \( \triangle DEF \). The sides are not equal, so the triangles are not congruent.
In simple words: Same shape, different size.

Teacher's Note:
1. Similar triangles are congruent only when the ratio of sides is 1.

Question 5. If \( \triangle ABC \sim \triangle PQR \) with \( \dfrac{BC}{QR} = \dfrac{1}{3} \), then find \( \dfrac{ar(\triangle PRQ)}{ar(\triangle BCA)} \).
Answer:
The ratio of areas equals the square of the ratio of corresponding sides: \( \dfrac{ar(\triangle PQR)}{ar(\triangle ABC)} = \left(\dfrac{QR}{BC}\right)^2 = 3^2 = \) 9.
In simple words: Area ratio is the square of the side ratio.

Teacher's Note:
1. Take the ratio the right way up: PQR over ABC.

Type II: Short Answer Type Questions (2 Marks Each)

Teacher's Note:
1. Check that the sides around the equal angle are in the same ratio.
2. The angle at P in triangle DPC is also 50°.

Question 7. If \( \triangle ABC \sim \triangle QRP \), \( \dfrac{ar(ABC)}{ar(PQR)} = \dfrac{9}{4} \), AB = 18 cm and BC = 15 cm, then find PR.
Answer:
\( \dfrac{ar(ABC)}{ar(QRP)} = \left(\dfrac{BC}{RP}\right)^2 = \dfrac{9}{4} \Rightarrow \dfrac{BC}{RP} = \dfrac{3}{2} \).
\( RP = \dfrac{2}{3} \times 15 = \) 10 cm.
In simple words: BC corresponds to RP.

Teacher's Note:
1. In ABC ~ QRP, side BC matches side RP, not PQ.

Question 8. If S is a point on side PQ of a \( \triangle PQR \) such that PS = QS = RS, then prove that \( PR^2 + QR^2 = PQ^2 \).
Answer:
PS = RS gives \( \angle P = \angle PRS \), and QS = RS gives \( \angle Q = \angle QRS \).
In \( \triangle PQR \): \( \angle P + \angle Q + \angle R = 180^\circ \Rightarrow 2(\angle PRS + \angle QRS) = 180^\circ \Rightarrow \angle PRQ = 90^\circ \).
By the Pythagoras theorem, \( PR^2 + QR^2 = PQ^2 \).
In simple words: S is the midpoint of the hypotenuse, so the angle at R is a right angle.

Teacher's Note:
1. Equal sides give equal base angles.
2. The angle sum then forces 90°.

Question 9. If \( \triangle ABC \sim \triangle DFE \), \( \angle A = 30^\circ \), \( \angle C = 50^\circ \), AB = 5 cm, AC = 8 cm and DF = 7.5 cm, then find DE and \( \angle F \).
Answer:
Matching vertices: A to D, B to F, C to E. So \( \dfrac{AB}{DF} = \dfrac{AC}{DE} \Rightarrow \dfrac{5}{7.5} = \dfrac{8}{DE} \Rightarrow DE = \) 12 cm.
\( \angle F = \angle B = 180^\circ - 30^\circ - 50^\circ = \) 100°.
In simple words: Read the similarity statement letter by letter.

Teacher's Note:
1. In ABC ~ DFE, B corresponds to F.
2. Use the angle sum for the missing angle.

Question 10. Corresponding sides of two similar triangles are in the ratio 2 : 3. If the area of the smaller triangle is 48 cm², find the area of the larger triangle.
Answer:
\( \dfrac{48}{\text{Area}} = \left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9} \Rightarrow \) Area \( = 48 \times \dfrac{9}{4} = \) 108 cm².
In simple words: Square the side ratio to get the area ratio.

Teacher's Note:
1. Area ratio 4 : 9 for side ratio 2 : 3.

Type III: Long Answer Type Questions (4 Marks Each)

Teacher's Note:
1. If a line divides two sides in the same ratio, it is parallel to the third side.
2. The x² terms cancel.

Teacher's Note:
1. Congruent triangles give equal corresponding parts (CPCT).
2. A line that divides two sides proportionally is parallel to the third side.

Question 13. Find the altitude of an equilateral triangle of side 8 cm.
Answer:
The altitude bisects the base, so \( h^2 = 8^2 - 4^2 = 48 \).
\( h = \) \( 4\sqrt{3} \) cm.
In simple words: Use h = (√3/2) × side.

Teacher's Note:
1. The altitude of an equilateral triangle of side a is \( \frac{\sqrt{3}}{2}a \).

Question 14. In a \( \triangle PQR \), \( PR^2 - PQ^2 = QR^2 \) and M is a point on side PR such that \( QM \perp PR \). Prove that \( QM^2 = PM \times MR \).
Answer:
\( PR^2 = PQ^2 + QR^2 \), so \( \angle PQR = 90^\circ \) (converse of the Pythagoras theorem).
QM is the perpendicular from the right angle to the hypotenuse, so \( \triangle QMP \sim \triangle RMQ \) (AA).
\( \dfrac{QM}{RM} = \dfrac{PM}{QM} \Rightarrow QM^2 = PM \times MR \).
In simple words: The altitude to the hypotenuse is the mean proportional of the two parts.

Teacher's Note:
1. First show the angle at Q is a right angle.
2. Then split the triangle by the altitude.

Question 15. ABCD is a trapezium in which \( AB \parallel DC \), and P and Q are points on AD and BC respectively such that \( PQ \parallel DC \). If PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD.
Answer:
A line parallel to the parallel sides divides the legs in the same ratio: \( \dfrac{AP}{PD} = \dfrac{BQ}{QC} \).
\( AP = \dfrac{35}{15} \times 18 = 42 \) cm, so \( AD = AP + PD = 42 + 18 = \) 60 cm.
In simple words: The parallel line cuts both legs in the ratio 35 : 15.

Teacher's Note:
1. Join AC (or extend the legs) to apply the basic proportionality theorem twice.

Question 16. The hypotenuse of a right triangle is 25 cm and, of the remaining two sides, one is longer than the other by 5 cm. Find the lengths of the other two sides.
Answer:
Let the shorter side be x cm. \( x^2 + (x + 5)^2 = 625 \Rightarrow x^2 + 5x - 300 = 0 \Rightarrow (x + 20)(x - 15) = 0 \).
x = 15 (x = -20 is rejected). The sides are 15 cm and 20 cm.
In simple words: A 3-4-5 triangle scaled by 5.

Teacher's Note:
1. Reject the negative root, a length cannot be negative.

Question 17. A street light bulb is fixed on a pole 6 m above the level of the street. If a woman of height 1.5 m casts a shadow of 3 m, find how far she is from the base of the pole.
Answer:
Let she stand d m from the base. The bulb, her head and the tip of the shadow are in a line, so \( \dfrac{6}{d + 3} = \dfrac{1.5}{3} \) (similar triangles).
\( d + 3 = 12 \Rightarrow d = \) 9 m.
In simple words: The pole and the woman form two similar right triangles.

Teacher's Note:
1. The pole and the woman are both vertical, so the triangles share the angle at the shadow tip.

Type IV: Very Long Answer Type Questions (5 Marks Each)

Teacher's Note:
1. Opposite sides of a parallelogram are parallel and equal.
2. Use the converse of the basic proportionality theorem in the big triangle AOC.

Teacher's Note:
1. The segments on l are in the ratio 6 : 9 : 12.
2. Split the length 36 cm in the same ratio.

Teacher's Note:
1. BD = BE because OB bisects DE.
2. Divide by OA·OB·OC at the end.

Question 21. The legs (sides other than the hypotenuse) of a right triangle are of lengths 16 cm and 8 cm. Find the length of the side of the largest square that can be inscribed in the triangle.
Answer:
The largest square has one corner at the right angle and its sides along the legs. Let its side be s; the remaining part of the leg 8 gives a small triangle similar to the big one: \( \dfrac{s}{16} = \dfrac{8 - s}{8} \).
\( 8s = 16(8 - s) \Rightarrow 24s = 128 \Rightarrow s = \) \( \dfrac{16}{3} \) cm.
In simple words: The square at the right angle gives s = ab/(a + b).

Teacher's Note:
1. The square with a corner at the right angle is larger than the one on the hypotenuse.
2. The small triangle cut off is similar to the whole triangle.