CBSE Class 10 Mathematics Triangles Assignment Set 03

Exam Tip: Remember that tangents drawn from an external point to a circle are equal, and the radius is perpendicular to the sides at the points of contact.

Question 2. ABC is a triangle. PQ is the line segment intersecting AB in P and AC in Q such that PQ parallel to BC and divides triangle ABC into two parts equal in area. Find BP:AB.
Answer: Since \(PQ \parallel BC\), triangle \(APQ\) is similar to triangle \(ABC\) (\(\triangle APQ \sim \triangle ABC\)).
The ratio of the area of triangle \(APQ\) to the area of triangle \(ABC\) is given by:
\(\frac{\text{Area}(APQ)}{\text{Area}(ABC)} = \frac{1}{2}\)
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides:
\(\left(\frac{AP}{AB}\right)^2 = \frac{\text{Area}(APQ)}{\text{Area}(ABC)} = \frac{1}{2}\)
\(\frac{AP}{AB} = \frac{1}{\sqrt{2}}\)
To find \(BP:AB\):
\(\frac{BP}{AB} = \frac{AB - AP}{AB} = 1 - \frac{AP}{AB} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}}\)
In simple words: Use the property that the ratio of areas of similar triangles is equal to the square of the ratio of their corresponding sides.

Exam Tip: Whenever a line divides a triangle into two equal areas, apply the area-similarity theorem for triangles.

Question 3. In a right triangle ABC, right angled at C, P and Q are points of the sides CA and CB respectively, which divide these sides in the ratio 2:1. Prove that:
(i) \(9AQ^2 = 9AC^2 + 4BC^2\)
(ii) \(9BP^2 = 9BC^2 + 4AC^2\)
(iii) \(9(AQ^2 + BP^2) = 13AB^2\)
Answer: (i) Since \(P\) divides \(CA\) in the ratio \(2:1\) from \(A\), we have \(CP = \frac{1}{3}AC\) and \(AP = \frac{2}{3}AC\). Similarly, \(CQ = \frac{2}{3}BC\) and \(BQ = \frac{1}{3}BC\).
In right-angled triangle \(ACQ\), applying Pythagoras theorem:
\(AQ^2 = AC^2 + CQ^2\)
\(AQ^2 = AC^2 + \left(\frac{2}{3}BC\right)^2\)
\(AQ^2 = AC^2 + \frac{4}{9}BC^2\)
Multiply both sides by 9:
\(9AQ^2 = 9AC^2 + 4BC^2\) --- (1)
(ii) In right-angled triangle \(BCP\), applying Pythagoras theorem:
\(BP^2 = BC^2 + CP^2\)
\(BP^2 = BC^2 + \left(\frac{2}{3}AC\right)^2\)
\(BP^2 = BC^2 + \frac{4}{9}AC^2\)
Multiply both sides by 9:
\(9BP^2 = 9BC^2 + 4AC^2\) --- (2)
(iii) Adding equations (1) and (2):
\(9(AQ^2 + BP^2) = (9AC^2 + 4BC^2) + (9BC^2 + 4AC^2)\)
\(9(AQ^2 + BP^2) = 13AC^2 + 13BC^2\)
\(9(AQ^2 + BP^2) = 13(AC^2 + BC^2)\)
Since triangle \(ABC\) is right-angled at \(C\), \(AC^2 + BC^2 = AB^2\).
Therefore, \(9(AQ^2 + BP^2) = 13AB^2\).
In simple words: Apply Pythagoras theorem separately to the right-angled triangles containing the given segments and add the resulting equations.

Exam Tip: Carefully write down the segment lengths given by the ratio theorem before applying Pythagoras theorem.

Question 4. P and Q are the mid points on the sides CA and CB respectively of triangle ABC right angled at C. Prove that \(4(AQ^2 + BP^2) = 5AB^2\)
Answer: In right-angled triangle \(ACQ\):
\(AQ^2 = AC^2 + CQ^2\)
Since \(Q\) is the midpoint of \(BC\), \(CQ = \frac{1}{2}BC\).
\(AQ^2 = AC^2 + \left(\frac{1}{2}BC\right)^2 = AC^2 + \frac{BC^2}{4}\)
\(4AQ^2 = 4AC^2 + BC^2\) --- (1)
In right-angled triangle \(BCP\):
\(BP^2 = BC^2 + CP^2\)
Since \(P\) is the midpoint of \(AC\), \(CP = \frac{1}{2}AC\).
\(BP^2 = BC^2 + \left(\frac{1}{2}AC\right)^2 = BC^2 + \frac{AC^2}{4}\)
\(4BP^2 = 4BC^2 + AC^2\) --- (2)
Adding equations (1) and (2):
\(4(AQ^2 + BP^2) = 5AC^2 + 5BC^2\)
\(4(AQ^2 + BP^2) = 5(AC^2 + BC^2)\)
Since \(AC^2 + BC^2 = AB^2\) (by Pythagoras theorem in \(\triangle ABC\)):
\(4(AQ^2 + BP^2) = 5AB^2\).
In simple words: Substitute midpoint values into Pythagoras theorem for triangles \(ACQ\) and \(BCP\), then add both equations.

Exam Tip: Midpoints divide lengths by 2, so squares of midpoints introduce a factor of \(\frac{1}{4}\).

Question 5. In an equilateral triangle ABC, the side BC is triseded at D. Prove that \(9AD^2 = 7AB^2\)
Answer: Let the side of the equilateral triangle be \(a\), so \(AB = BC = CA = a\).
Draw an altitude \(AE\) perpendicular to \(BC\). Since \(\triangle ABC\) is equilateral, \(E\) is the midpoint of \(BC\), so \(BE = EC = \frac{a}{2}\).
Since \(BC\) is trisected at \(D\), \(BD = \frac{1}{3}a\).
Therefore, \(DE = BE - BD = \frac{a}{2} - \frac{a}{3} = \frac{a}{6}\).
In right-angled triangle \(ABE\):
\(AE^2 = AB^2 - BE^2 = a^2 - \left(\frac{a}{2}\right)^2 = \frac{3}{4}a^2\)
In right-angled triangle \(ADE\), applying Pythagoras theorem:
\(AD^2 = AE^2 + DE^2\)
\(AD^2 = \frac{3}{4}a^2 + \left(\frac{a}{6}\right)^2\)
\(AD^2 = \frac{3}{4}a^2 + \frac{a^2}{36} = \frac{27a^2 + a^2}{36} = \frac{28}{36}a^2 = \frac{7}{9}a^2\)
Since \(a = AB\):
\(AD^2 = \frac{7}{9}AB^2\)
\(9AD^2 = 7AB^2\).
In simple words: Drop an altitude to form right-angled triangles, express the distances in terms of side \(a\), and apply Pythagoras theorem.

Exam Tip: Constructing an altitude is the standard technique for solving problems involving equilateral triangles and non-vertex points.

Question 6. There is a staircase as shown in figure connecting points A and B. Measurements of steps are marked in the figure. Find the straight distance between A and B.
Answer: The staircase consists of horizontal runs and vertical rises. Let us sum all the horizontal lengths and vertical heights between points \(A\) and \(B\).
Total horizontal distance (\(x\)) = \(2 + 2 + 2 + 2 = 8\)
Total vertical distance (\(y\)) = \(1 + 1.6 + 1.6 + 1.8 = 6\)
The straight distance between \(A\) and \(B\) forms the hypotenuse of a right-angled triangle with base \(8\) and height \(6\).
\(AB^2 = (\text{Total Horizontal})^2 + (\text{Total Vertical})^2\)
\(AB^2 = 8^2 + 6^2 = 64 + 36 = 100\)
\(AB = \sqrt{100} = 10\text{ units}\).
In simple words: Combine all small horizontal steps into a single base and all vertical steps into a single height, then use Pythagoras theorem.

Exam Tip: A stepped path can always be treated as a single large right-angled triangle by adding up individual components.

Question 7. Find the length of the second diagonal of a rhombus, whose side is 5cm and one of the diagonals is 6cm.
Answer: Let the diagonals of the rhombus intersect at \(O\) at right angles (\(90^\circ\)).
Let diagonal \(d_1 = 6\text{ cm}\). The diagonals of a rhombus bisect each other at right angles.
Half of the first diagonal = \(\frac{6}{2} = 3\text{ cm}\).
The side of the rhombus is \(5\text{ cm}\), which acts as the hypotenuse of the right-angled triangle formed by half-diagonals.
Using Pythagoras theorem:
\(\text{Side}^2 = (\text{Half of } d_1)^2 + (\text{Half of } d_2)^2\)
\(5^2 = 3^2 + BO^2\)
\(25 = 9 + BO^2\)
\(BO^2 = 25 - 9 = 16\)
\(BO = 4\text{ cm}\)
Therefore, the full length of the second diagonal \(d_2 = 2 \times BO = 2 \times 4 = 8\text{ cm}\).
In simple words: Diagonals of a rhombus bisect each other at \(90^\circ\). Use Pythagoras theorem on one of the four right triangles formed inside the rhombus.

Exam Tip: Always remember that the diagonals of a rhombus bisect each other perpendicularly.

Question 8. Prove that three times the sum of the squares of the sides of a triangle is equal to four times the sum of the squares of the medias of the triangle.
Answer: To prove: \(3(AB^2 + BC^2 + CA^2) = 4(AD^2 + BE^2 + CF^2)\)
In any triangle, using the Apollonius theorem (or property of medians):
For median \(AD\) on side \(BC\):
\(AB^2 + AC^2 = 2\left(AD^2 + \frac{BC^2}{4}\right)\)
\(2(AB^2 + AC^2) = 4AD^2 + BC^2\) --- (1)
Similarly, for median \(BE\) on side \(AC\):
\(2(AB^2 + BC^2) = 4BE^2 + AC^2\) --- (2)
And for median \(CF\) on side \(AB\):
\(2(AC^2 + BC^2) = 4CF^2 + AB^2\) --- (3)
Adding equations (1), (2), and (3):
\(2(2AB^2 + 2BC^2 + 2AC^2) = 4(AD^2 + BE^2 + CF^2) + AB^2 + BC^2 + AC^2\)
\(4(AB^2 + BC^2 + AC^2) - (AB^2 + BC^2 + AC^2) = 4(AD^2 + BE^2 + CF^2)\)
\(3(AB^2 + BC^2 + AC^2) = 4(AD^2 + BE^2 + CF^2)\).
In simple words: Apply the median formula (Appollonius theorem) for all three medians and add the resulting equations together.

Exam Tip: The relation between sides and medians of a triangle is a standard identity derived from the theorem of Apollonius.

Question 9. ABC is an isosceles triangle in which AB=AC=10cm, BC=12. PQRS is a rectangle inside the isosceles triangle. Given PQ=SR=y cm, PS=QR=2x. Prove that \(x = 6 - \frac{3y}{4}\)
Answer: Let \(AL\) be the altitude from \(A\) to \(BC\). Since \(ABC\) is an isosceles triangle with \(AB = AC\), the altitude \(AL\) bisects the base \(BC\).
\(BL = \frac{BC}{2} = \frac{12}{2} = 6\text{ cm}\).
In right-angled triangle \(ABL\), using Pythagoras theorem:
\(AL^2 = AB^2 - BL^2 = 10^2 - 6^2 = 100 - 36 = 64\)
\(AL = 8\text{ cm}\).
Since \(PQRS\) is a rectangle and \(PQ \parallel AB\), triangle \(BPQ\) is similar to triangle \(BAL\) (\(\triangle BPQ \sim \triangle BAL\)).
Using the properties of similar triangles, the ratio of corresponding sides is equal:
\(\frac{PQ}{AL} = \frac{BP}{BL}\)
From the figure, \(PS = 2x\), so \(BP = BL - PS/2 = 6 - x\) (since \(P\) is symmetrically placed on \(BL\)).
Substitute \(PQ = y\), \(AL = 8\), \(BP = 6 - x\), and \(BL = 6\):
\(\frac{y}{8} = \frac{6 - x}{6}\)
Simplify the equation:
\(\frac{6 - x}{y} = \frac{6}{8} = \frac{3}{4}\)
\(6 - x = \frac{3}{4}y\)
\(x = 6 - \frac{3}{4}y\).
Hence proved.
In simple words: Find the height of the triangle using Pythagoras theorem, then use the similarity between triangle \(BPQ\) and triangle \(BAL\).

Exam Tip: Properties of similar triangles are extremely helpful when geometric shapes like rectangles are inscribed inside triangles.

Question 10. If ABC is an obtuse angled triangle, obtuse angled at B and if \(AD \perp CB\) Prove that \(AC^2 = AB^2 + BC^2 + 2BC \times BD\)
Answer: In right-angled triangle \(ADC\), applying Pythagoras theorem:
\(AC^2 = AD^2 + CD^2\)
Since \(CD = BC + BD\), substitute this into the equation:
\(AC^2 = AD^2 + (BC + BD)^2\)
Expand the binomial:\((BC + BD)^2 = BC^2 + BD^2 + 2BC \times BD\)
\(AC^2 = AD^2 + BC^2 + BD^2 + 2BC \times BD\)
In right-angled triangle \(ABD\), \(AB^2 = AD^2 + BD^2\). Substituting this:
\(AC^2 = (AD^2 + BD^2) + BC^2 + 2BC \times BD\)
\(AC^2 = AB^2 + BC^2 + 2BC \times BD\).
Hence proved.
In simple words: Express \(CD\) as \(BC + BD\) and apply Pythagoras theorem to the outer and inner right-angled triangles.

Exam Tip: This is the generalized Pythagoras theorem for obtuse-angled triangles.

Question 11. If ABC is an acute angled triangle, acute angled at B and \(AD \perp BC\) prove that \(AC^2 = AB^2 + BC^2 - 2BC \times BD\)
Answer: In right-angled triangle \(ADC\), applying Pythagoras theorem:
\(AC^2 = AD^2 + CD^2\)
Since \(CD = BC - BD\), substitute this into the equation:
\(AC^2 = AD^2 + (BC - BD)^2\)
Expand the binomial:\((BC - BD)^2 = BC^2 + BD^2 - 2BC \times BD\)
\(AC^2 = AD^2 + BC^2 + BD^2 - 2BC \times BD\)
In right-angled triangle \(ABD\), \(AB^2 = AD^2 + BD^2\). Substituting this:
\(AC^2 = (AD^2 + BD^2) + BC^2 - 2BC \times BD\)
\(AC^2 = AB^2 + BC^2 - 2BC \times BD\).
Hence proved.
In simple words: Substitute \(CD = BC - BD\) into Pythagoras theorem for triangle \(ADC\) and simplify.

Exam Tip: This is the generalized Pythagoras theorem (Appollonius / Law of Cosines form) for acute-angled triangles.

Question 12. Prove that in any triangle the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median, which bisects the third side.
Answer: To prove: \(AB^2 + AC^2 = 2AD^2 + 2\left(\frac{1}{2}BC\right)^2\)
Draw \(AE \perp BC\). Let \(AD\) be the median, so \(BD = DC = \frac{1}{2}BC\).
In triangle \(ABD\), since \(\angle D\) can be obtuse or acute, using the generalized Pythagoras theorem (Questions 10 and 11):
\(AB^2 = AD^2 + BD^2 + 2BD \times DE\) --- (for obtuse/acute components, or by combining right triangles \(ADE\) and \(ABE\))
More simply, in right triangle \(ABE\):
\(AB^2 = AE^2 + BE^2 = AE^2 + (BD - DE)^2 = AE^2 + BD^2 + DE^2 - 2BD \cdot DE\)
And in right triangle \(ADC\):
\(AC^2 = AE^2 + EC^2 = AE^2 + (DC + DE)^2 = AE^2 + DC^2 + DE^2 + 2DC \cdot DE\)
Since \(BD = DC\), adding both equations gives:
\(AB^2 + AC^2 = 2(AE^2 + DE^2) + 2BD^2\)
Since \(AE^2 + DE^2 = AD^2\) and \(BD = \frac{1}{2}BC\):
\(AB^2 + AC^2 = 2AD^2 + 2\left(\frac{1}{2}BC\right)^2\).
Hence proved.
In simple words: Drop a perpendicular from vertex \(A\) to base \(BC\), apply Pythagoras theorem on triangles \(ABE\) and \(ACE\), and combine them using median properties.

Exam Tip: This is the statement and proof of Apollonius theorem.

Question 13. If A be the area of a right triangle and b one of the sides containing the right angle, prove that the length of the altitude on the hypotenuse is \(\frac{2Ab}{\sqrt{b^4 + 4A^2}}\)
Answer: Let the sides containing the right angle be \(b\) and \(q\).
The area of the right triangle is given by:
\(A = \frac{1}{2} \times b \times q\)
\(q = \frac{2A}{b}\) --- (1)
The hypotenuse \(PR\) can be found using Pythagoras theorem:
\(PR = \sqrt{b^2 + q^2} = \sqrt{b^2 + \left(\frac{2A}{b}\right)^2} = \sqrt{b^2 + \frac{4A^2}{b^2}} = \frac{\sqrt{b^4 + 4A^2}}{b}\)
Let the altitude on the hypotenuse be \(NQ\). By equating the two ways of calculating the area of triangle \(PQR\):
\(\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}\)
\(A = \frac{1}{2} \times PR \times NQ\)
\(NQ = \frac{2A}{PR}\)
Substitute the value of \(PR\):
\(NQ = \frac{2A}{\frac{\sqrt{b^4 + 4A^2}}{b}} = \frac{2Ab}{\sqrt{b^4 + 4A^2}}\).
Hence proved.
In simple words: Express the second side in terms of area and \(b\), find the hypotenuse, and then use the area formula with the hypotenuse as base to find the altitude.

Exam Tip: Area of a triangle can be calculated in multiple ways using different sides as bases; equating them is a powerful problem-solving tool.

Question 14. ABC is a right triangle right-angled at C and \(AC = \sqrt{3}BC\). Prove that \(\angle ABC = 60^\circ\).
Answer: In right-angled triangle \(ABC\) at \(C\), the trigonometric ratio for \(\tan B\) is:
\(\tan B = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AC}{BC}\)
Substitute \(AC = \sqrt{3}BC\):
\(\tan B = \frac{\sqrt{3}BC}{BC} = \sqrt{3}\)
We know that \(\tan 60^\circ = \sqrt{3}\).
Therefore, \(B = 60^\circ\), which means \(\angle ABC = 60^\circ\).
Hence proved.
In simple words: Use the definition of tangent in a right-angled triangle and standard trigonometric table values.

Exam Tip: Memorize standard trigonometric ratios for angles like \(30^\circ\), \(45^\circ\), and \(60^\circ\).

Question 15. ABCD is a rectangle. \(\triangle ADE\) and \(\triangle ABF\) are two triangles such that \(\angle E = \angle F\) as shown in the figure. Prove that \(AD \times AF = AE \times AB\).
Answer: Consider triangle \(ADE\) and triangle \(ABF\):
\(\angle D = \angle B = 90^\circ\) (since \(ABCD\) is a rectangle)
\(\angle E = \angle F\) (given)
By AA (Angle-Angle) similarity criterion, \(\triangle ADE \sim \triangle ABF\).
Since the triangles are similar, the ratio of their corresponding sides is equal:
\(\frac{AD}{AB} = \frac{AE}{AF}\)
Cross-multiplying gives:
\(AD \times AF = AE \times AB\).
Hence proved.
In simple words: Prove the two triangles are similar using AA similarity, then use the proportional property of corresponding sides.

Exam Tip: Whenever you need to prove a product relation between lengths (like \(AD \times AF = AE \times AB\)), look for similar triangles and set up ratios.

Question 16. In the given figure, \(\angle AEF = \angle AFE\) and E is the mid-point of CA. Prove that \(\frac{BD}{CD} = \frac{BF}{CE}\)
Answer: Construction: Draw \(CG \parallel DF\).
In triangle \(BDF\), since \(CG \parallel DF\), by Basic Proportionality Theorem (BPT):
\(\frac{BD}{CD} = \frac{BF}{GF}\) --- (1)
In triangle \(AFE\), we are given \(\angle AEF = \angle AFE\).
Therefore, opposite sides are equal: \(AF = AE\).
Since \(E\) is the midpoint of \(CA\), \(AE = EC\).
Thus, \(AF = AE = EC\) --- (2)
In triangle \(ACG\), since \(E\) is the midpoint of \(CA\) and \(EF \parallel CG\), by midpoint theorem or proportionality, \(F\) is the midpoint of \(AG\), which gives \(GF = AF\).
Substituting \(GF = CE\) into equation (1):
\(\frac{BD}{CD} = \frac{BF}{CE}\).
Hence proved.
In simple words: Use a parallel line construction to apply the Basic Proportionality Theorem, and combine it with the properties of isosceles triangles and midpoints.

Exam Tip: Auxiliary parallel lines are crucial for connecting ratios of line segments in triangle geometry.

Question 17. ABCD is a parallelogram in the given figure, AB is divided at P and CD and Q so that \(AP:PB = 3:2\) and \(CQ:QD = 4:1\). If PQ meets AC at R, prove that \(AR = \frac{3}{7}AC\).
Answer: Consider triangle \(APR\) and triangle \(CQR\):
\(\angle PAR = \angle QCR\) (alternate interior angles since \(AB \parallel CD\))
\(\angle ARP = \angle CRQ\) (vertically opposite angles)
By AA similarity criterion, \(\triangle APR \sim \triangle CQR\).
Therefore, the ratio of corresponding sides is equal:
\(\frac{AP}{CQ} = \frac{PR}{QR} = \frac{AR}{CR}\)
Given \(AP:PB = 3:2\), so \(AP = \frac{3}{5}AB\).
Given \(CQ:QD = 4:1\), so \(CQ = \frac{4}{5}CD\).
Since \(ABCD\) is a parallelogram, \(AB = CD\).
\(\frac{AP}{CQ} = \frac{\frac{3}{5}AB}{\frac{4}{5}AB} = \frac{3}{4}\)
Thus, \(\frac{AR}{CR} = \frac{3}{4}\), which means \(CR = \frac{4}{3}AR\).
Now, \(AC = AR + CR = AR + \frac{4}{3}AR = \frac{7}{3}AR\).
Rearranging for \(AR\):
\(AR = \frac{3}{7}AC\).
Hence proved.
In simple words: Establish triangle similarity between \(APR\) and \(CQR\), use the ratio of side lengths from the given divisions, and express \(AR\) as a fraction of total diagonal \(AC\).

Exam Tip: Parallel sides in parallelograms create alternate interior angles, which are essential for proving triangle similarity.

Question 18. Prove that the area of a rhombus on the hypotenuse of a right-angled triangle, with one of the angles as \(60^\circ\), is equal to the sum of the areas of rhombuses with one of their angles as \(60^\circ\) drawn on the other two sides.
Answer: Let the sides of the right-angled triangle be \(a\), \(b\), and \(c\), where \(c\) is the hypotenuse, so \(a^2 + b^2 = c^2\).
The area of a rhombus of side \(x\) and acute angle \(60^\circ\) is given by:
\(\text{Area} = x^2 \times \sin(60^\circ) = \frac{\sqrt{3}}{2}x^2\)
The area of the rhombus on side \(a\) is \(\frac{\sqrt{3}}{2}a^2\).
The area of the rhombus on side \(b\) is \(\frac{\sqrt{3}}{2}b^2\).
The area of the rhombus on hypotenuse \(c\) is \(\frac{\sqrt{3}}{2}c^2\).
Since \(a^2 + b^2 = c^2\), multiplying both sides by \(\frac{\sqrt{3}}{2}\) gives:
\(\frac{\sqrt{3}}{2}a^2 + \frac{\sqrt{3}}{2}b^2 = \frac{\sqrt{3}}{2}c^2\)
Therefore, the sum of the areas of the rhombuses on the two smaller sides equals the area of the rhombus on the hypotenuse.
Hence proved.
In simple words: Express the area of a rhombus in terms of its side and angle, then use Pythagoras theorem to show the sum of areas equals the area on the hypotenuse.

Exam Tip: Similar geometric figures (like squares, equilateral triangles, or rhombuses with the same angle) constructed on the sides of a right triangle satisfy Pythagoras theorem relations.

Question 19. An aeroplane leaves an airport and flies due north at a speed of \(1000\text{ km/h}\). At the same time, another plane leaves the same airport and flies due west at a speed of \(1200\text{ km/h}\). How far apart will the two planes after \(1\frac{1}{2}\) hours? (Ans: \(300\sqrt{61}\text{ Km}\))
Answer: Time elapsed (\(t\)) = \(1\frac{1}{2}\text{ hours} = 1.5\text{ hours}\).
Distance travelled by the plane flying due North (\(ON\)) = \(\text{Speed} \times \text{Time} = 1000 \times 1.5 = 1500\text{ km}\).
Distance travelled by the plane flying due West (\(OW\)) = \(\text{Speed} \times \text{Time} = 1200 \times 1.5 = 1800\text{ km}\).
Since North and West directions are perpendicular to each other (\(90^\circ\)), the distance between the two planes forms the hypotenuse (\(NW\)) of a right-angled triangle.
Using Pythagoras theorem:
\(NW^2 = ON^2 + OW^2\)
\(NW^2 = 1500^2 + 1800^2 = 2250000 + 3240000 = 5490000\)
\(NW = \sqrt{5490000} = \sqrt{90000 \times 61} = 300\sqrt{61}\text{ km}\).
In simple words: Calculate the distance covered by each plane by multiplying speed and time, then use Pythagoras theorem since North and West are perpendicular directions.

Exam Tip: Always calculate distances first (\(\text{Distance} = \text{Speed} \times \text{Time}\)) before applying Pythagoras theorem.

Question 20. ABC is a right-angled isosceles triangle, right-angled at B. AP, the bisector of \(\angle BAC\), intersects BC at P. Prove that \(AC^2 = AP^2 + 2(1+\sqrt{2})BP^2\)
Answer: In the right-angled isosceles triangle \(ABC\) right-angled at \(B\), \(AB = BC\).
Using Pythagoras theorem, \(AC^2 = AB^2 + BC^2 = 2AB^2\), so \(AC = \sqrt{2}AB\).
By the Angle Bisector Theorem in triangle \(ABC\) where \(AP\) is the bisector of \(\angle A\):
\(\frac{AB}{AC} = \frac{BP}{CP}\)
Substitute \(AC = \sqrt{2}AB\):
\(\frac{AB}{\sqrt{2}AB} = \frac{BP}{CP}\)
\(CP = \sqrt{2}BP\)
Since \(BC = BP + CP\), we have \(BC = BP + \sqrt{2}BP = (1 + \sqrt{2})BP\).
Since \(AB = BC\), \(AB = (1 + \sqrt{2})BP\).
In right-angled triangle \(ABP\), applying Pythagoras theorem:
\(AP^2 = AB^2 + BP^2\)
Substitute \(AB = (1 + \sqrt{2})BP\):
\(AP^2 = \left((1 + \sqrt{2})BP\right)^2 + BP^2 = (1 + 2 + 2\sqrt{2})BP^2 + BP^2 = (4 + 2\sqrt{2})BP^2\)
Also, from \(AC^2 = 2AB^2\), substitute \(AB = (1 + \sqrt{2})BP\):
\(AC^2 = 2\left((1 + \sqrt{2})BP\right)^2 = 2(3 + 2\sqrt{2})BP^2 = (6 + 4\sqrt{2})BP^2\)
Simplifying both sides shows that \(AC^2 = AP^2 + 2(1 + \sqrt{2})BP^2\).
Hence proved.
In simple words: Use the angle bisector theorem to relate side segments, then apply Pythagoras theorem in triangle \(ABP\).

Exam Tip: Angle bisector theorem states that the bisector of an angle divides the opposite side in the ratio of the adjacent sides.