Table of Contents
Exercise 8.4 Solutions
Class X Mathematics textbook by the State Council of Educational Research and Training, Telangana, Hyderabad
Problem 1
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Solution:
Let ABCD be a rhombus with side length ‘a’ and diagonals d₁ and d₂ intersecting at O.
Properties of rhombus:
1. All sides equal: AB = BC = CD = DA = a
2. Diagonals bisect each other at right angles: AO = OC = d₁/2, BO = OD = d₂/2
Using Pythagoras theorem in ΔAOB:
\(AO^2 + BO^2 = AB^2\) ⇒ \(\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = a^2\)
\(\Rightarrow \frac{d_1^2}{4} + \frac{d_2^2}{4} = a^2\) ⇒ \(d_1^2 + d_2^2 = 4a^2\)
Sum of squares of all sides = \(4a^2\)
Thus, \(AB^2 + BC^2 + CD^2 + DA^2 = d_1^2 + d_2^2\)
Problem 2
ABC is a right triangle right angled at B. Let D and E be any points on AB and BC respectively. Prove that \( \text{AE}^2 + \text{CD}^2 = \text{AC}^2 + \text{DE}^2 \).
Solution:
Using Pythagoras theorem in various triangles:
In ΔABE: \(AE^2 = AB^2 + BE^2\)
In ΔCBD: \(CD^2 = CB^2 + BD^2\)
Adding: \(AE^2 + CD^2 = (AB^2 + CB^2) + (BE^2 + BD^2)\)
But \(AB^2 + CB^2 = AC^2\) (from ΔABC)
And \(BE^2 + BD^2 = DE^2\) (from ΔDBE, since ∠DBE = 90°)
Thus, \(AE^2 + CD^2 = AC^2 + DE^2\)
Problem 3
Prove that three times the square of any side of an equilateral triangle is equal to four times the square of the altitude.
Solution:
Let ABC be equilateral triangle with side ‘a’ and height ‘h’.
The altitude divides the base into two equal parts of length a/2.
Using Pythagoras theorem in the right triangle formed by altitude:
\(h^2 + \left(\frac{a}{2}\right)^2 = a^2\)
\(\Rightarrow h^2 = a^2 – \frac{a^2}{4} = \frac{3a^2}{4}\)
\(\Rightarrow 4h^2 = 3a^2\)
Thus, \(3a^2 = 4h^2\)
Problem 4
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that \( \text{PM}^2 = \text{QM} \cdot \text{MR} \).
Solution:
In right ΔPQR, PM is the altitude to hypotenuse QR.
By the Right Triangle Altitude Theorem:
1. \(PM^2 = QM \cdot MR\)
2. \(PQ^2 = QM \cdot QR\)
3. \(PR^2 = MR \cdot QR\)
Thus, the first property directly gives \(PM^2 = QM \cdot MR\)
Problem 5
ABD is a triangle right angled at A and AC ⊥ BD. Show that:
(i) \( \text{AB}^2 = \text{BC} \cdot \text{BD} \)
(ii) \( \text{AC}^2 = \text{BC} \cdot \text{DC} \)
(iii) \( \text{AD}^2 = \text{BD} \cdot \text{CD} \)
Solution:
This is similar to the right triangle altitude theorem.
(i) In ΔABC and ΔDBA:
∠B is common, ∠BAC = ∠BAD = 90° ⇒ ΔABC ∼ ΔDBA by AA
Thus, \(\frac{AB}{DB} = \frac{BC}{BA}\) ⇒ \(AB^2 = BC \cdot BD\)
(ii) In ΔABC and ΔDAC:
∠ACB = ∠DCA, both right angles ⇒ ΔABC ∼ ΔDAC by AA
Thus, \(\frac{AC}{DC} = \frac{BC}{AC}\) ⇒ \(AC^2 = BC \cdot DC\)
(iii) In ΔACD and ΔBAD:
∠D is common, ∠ACD = ∠BAD = 90° ⇒ ΔACD ∼ ΔBAD by AA
Thus, \(\frac{AD}{BD} = \frac{CD}{AD}\) ⇒ \(AD^2 = BD \cdot CD\)
Problem 6
ABC is an isosceles triangle right angled at C. Prove that \( \text{AB}^2 = 2\text{AC}^2 \).
Solution:
Given: AC = BC (isosceles), ∠C = 90°
By Pythagoras theorem:
\(AB^2 = AC^2 + BC^2 = AC^2 + AC^2 = 2AC^2\)
Thus, \(AB^2 = 2AC^2\)
Problem 7
‘O’ is any point in the interior of a triangle ABC. If OD ⊥ BC, OE ⊥ AC and OF ⊥ AB, show that:
(i) \( \text{OA}^2 + \text{OB}^2 + \text{OC}^2 – \text{OD}^2 – \text{OE}^2 – \text{OF}^2 = \text{AF}^2 + \text{BD}^2 + \text{CE}^2 \)
(ii) \( \text{AF}^2 + \text{BD}^2 + \text{CE}^2 = \text{AE}^2 + \text{CD}^2 + \text{BF}^2 \)
Solution:
(i) Using Pythagoras theorem in various right triangles:
In ΔAFO: \(AF^2 = OA^2 – OF^2\)
In ΔBDO: \(BD^2 = OB^2 – OD^2\)
In ΔCEO: \(CE^2 = OC^2 – OE^2\)
Adding: \(AF^2 + BD^2 + CE^2 = (OA^2 + OB^2 + OC^2) – (OF^2 + OD^2 + OE^2)\)
(ii) Similarly:
\(AE^2 = OA^2 – OE^2\), \(CD^2 = OC^2 – OD^2\), \(BF^2 = OB^2 – OF^2\)
Thus, \(AE^2 + CD^2 + BF^2 = (OA^2 + OB^2 + OC^2) – (OE^2 + OD^2 + OF^2)\)
Which equals \(AF^2 + BD^2 + CE^2\) from part (i)
Problem 8
A wire attached to a vertical pole of height 18m is 24m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution:
This forms a right triangle with:
Height = 18m (one leg), Hypotenuse = 24m (wire)
Let distance from pole = x (other leg)
By Pythagoras theorem: \(x^2 + 18^2 = 24^2\)
\(x^2 = 576 – 324 = 252\)
\(x = \sqrt{252} = 6\sqrt{7} \approx 15.87 \, \text{m}\)
Problem 9
Two poles of heights 6m and 11m stand on a plane ground. If the distance between the feet of the poles is 12m find the distance between their tops.
Solution:
Height difference = 11 – 6 = 5m
Horizontal distance = 12m
Distance between tops forms hypotenuse of right triangle:
\(d^2 = 5^2 + 12^2 = 25 + 144 = 169\)
\(d = \sqrt{169} = 13 \, \text{m}\)
Problem 10
In an equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\) BC. Prove that \(9AD^2 = 7AB^2\).
Solution:
Let AB = BC = CA = a, BD = a/3 ⇒ DC = 2a/3
Draw altitude AE from A to BC. In equilateral triangle, E is midpoint.
BE = a/2 ⇒ DE = BE – BD = a/2 – a/3 = a/6
AE = \(\frac{a\sqrt{3}}{2}\) (height of equilateral triangle)
In ΔADE: \(AD^2 = AE^2 + DE^2 = \frac{3a^2}{4} + \frac{a^2}{36} = \frac{28a^2}{36} = \frac{7a^2}{9}\)
Thus, \(9AD^2 = 7a^2 = 7AB^2\)
Problem 11
In the given figure, ABC is a triangle right angled at B. D and E are points on BC trisect it. Prove that \(8AE^2 = 3AC^2 + 5AD^2\).
Solution:
Let BC = 3a ⇒ BD = a, DE = a, EC = a
Let AB = c
Using Pythagoras theorem:
\(AD^2 = AB^2 + BD^2 = c^2 + a^2\)
\(AE^2 = AB^2 + BE^2 = c^2 + (2a)^2 = c^2 + 4a^2\)
\(AC^2 = AB^2 + BC^2 = c^2 + 9a^2\)
Now, \(3AC^2 + 5AD^2 = 3(c^2 + 9a^2) + 5(c^2 + a^2) = 8c^2 + 32a^2 = 8(c^2 + 4a^2) = 8AE^2\)
Problem 12
ABC is an isosceles triangle right angled at B. Similar triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of \(\triangle ABE\) and \(\triangle ACD\).
Solution:
Let AB = BC = a (isosceles right triangle)
Then AC = \(a\sqrt{2}\) (hypotenuse)
Since triangles are similar, ratio of areas = (ratio of corresponding sides)²
\(\frac{\text{Area } \triangle ABE}{\text{Area } \triangle ACD} = \left(\frac{AB}{AC}\right)^2 = \left(\frac{a}{a\sqrt{2}}\right)^2 = \frac{1}{2}\)
Problem 13
Equilateral triangles are drawn on the three sides of a right angled triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.
Solution:
Let right triangle have legs a, b and hypotenuse c (a² + b² = c²)
Area of equilateral triangle with side s = \(\frac{\sqrt{3}}{4}s^2\)
Area on hypotenuse = \(\frac{\sqrt{3}}{4}c^2\)
Sum of areas on legs = \(\frac{\sqrt{3}}{4}a^2 + \frac{\sqrt{3}}{4}b^2 = \frac{\sqrt{3}}{4}(a^2 + b^2) = \frac{\sqrt{3}}{4}c^2\)
Thus, area on hypotenuse = sum of areas on legs
Problem 14
Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangles described on its diagonal.
Solution:
Let square have side length ‘a’, then diagonal = \(a\sqrt{2}\)
Area of equilateral triangle on side = \(\frac{\sqrt{3}}{4}a^2\)
Area of equilateral triangle on diagonal = \(\frac{\sqrt{3}}{4}(a\sqrt{2})^2 = \frac{\sqrt{3}}{4}(2a^2) = \frac{\sqrt{3}}{2}a^2\)
Thus, area on side = ½ area on diagonal
