Exercise 16.1: Perimeter and Area of a Triangle
1. The hypotenuse of a right-angled triangle is $25\text{ m}$. If one of its remaining two sides is $\frac{3}{4}$ of the other, find the lengths of the two sides.
Solution: Let, the length of one side be $x\text{ m}$.
Then, the length of the other side $= \frac{3x}{4}\text{ m}$.
And the hypotenuse $= 25\text{ m}$.
According to Pythagoras’ theorem for a right-angled triangle, we get
$\text{(Base)}^2 + \text{(Perpendicular)}^2 = \text{(Hypotenuse)}^2$
$\text{or, } x^2 + \left(\frac{3x}{4}\right)^2 = 25^2$
$\text{or, } x^2 + \frac{9x^2}{16} = 625$
$\text{or, } \frac{16x^2 + 9x^2}{16} = 625$
$\text{or, } \frac{25x^2}{16} = 625$
$\text{or, } 25x^2 = 625 \times 16$
$\text{or, } x^2 = \frac{625 \times 16}{25}$
$\text{or, } x^2 = 25 \times 16$
$\text{or, } x^2 = 400$
$\text{or, } x = \sqrt{400}$
$\text{or, } x = 20$
Therefore, the length of one side is $20\text{ m}$.
And the length of the other side $= \frac{3 \times 20}{4} = 15\text{ m}$.
Answer:$15\text{ m}$ and $20\text{ m}$.
2. A ladder with length $20\text{ m}$ stands vertically against a wall. How much further should the lower end of the ladder be moved so that its upper end descends $4\text{ metre}$ ?
Solution: Given,
Length of the ladder (Hypotenuse) $= 20\text{ m}$.
Initial height of the wall $= 20\text{ m}$.
When the top end of the ladder moves down by $4\text{ m}$, the new height (Perpendicular) $= (20 – 4)\text{ m} = 16\text{ m}$.
Let the foot of the ladder be moved $x\text{ m}$ away from the wall ($\text{Base} = x\text{ m}$).
By Pythagoras’ theorem, we obtain
$\text{(Base)}^2 + \text{(Perpendicular)}^2 = \text{(Hypotenuse)}^2$
$\text{or, } x^2 + 16^2 = 20^2$
$\text{or, } x^2 + 256 = 400$
$\text{or, } x^2 = 400 – 256$
$\text{or, } x^2 = 144$
$\text{or, } x = \sqrt{144}$
$\text{or, } x = 12$
Answer: It should be pulled $12\text{ m}$ away from the wall.
3. The perimeter of an isosceles triangle is $16\text{ m}$. If the length of equal sides is $\frac{5}{6}$ of base, find the area of the triangle.
Solution: Let, the length of the base be $b = x\text{ m}$.
Therefore, the length of each equal side is $a = \frac{5x}{6}\text{ m}$.
According to the question,
$\text{Perimeter} = 16\text{ m}$
$\text{or, } a + a + b = 16$
$\text{or, } \frac{5x}{6} + \frac{5x}{6} + x = 16$
$\text{or, } \frac{5x + 5x + 6x}{6} = 16$
$\text{or, } \frac{16x}{6} = 16$
$\text{or, } 16x = 16 \times 6$
$\text{or, } x = 6$
Therefore, base length $b = 6\text{ m}$.
Length of each equal side $a = \frac{5 \times 6}{6} = 5\text{ m}$.
We know that the area of an isosceles triangle $= \frac{b}{4}\sqrt{4a^2 – b^2}\text{ sq. units}$
$= \frac{6}{4}\sqrt{4(5)^2 – 6^2}\text{ sq. m}$
$= \frac{3}{2}\sqrt{4(25) – 36}\text{ sq. m}$
$= \frac{3}{2}\sqrt{100 – 36}\text{ sq. m}$
$= \frac{3}{2}\sqrt{64}\text{ sq. m}$
$= \frac{3}{2} \times 8\text{ sq. m}$
$= 12\text{ sq. m}$
Answer:$12\text{ sq. m}$.
4. The lengths of the two sides of a triangle are $25\text{ cm}$ and $27\text{ cm}$, and perimeter is $84\text{ cm}$. Find the area of the triangle.
Solution: Let,
First side $a = 25\text{ cm}$
Second side $b = 27\text{ cm}$
Third side $= c\text{ cm}$
$\text{Perimeter} = 84\text{ cm}$
$\text{or, } a + b + c = 84$
$\text{or, } 25 + 27 + c = 84$
$\text{or, } 52 + c = 84$
$\text{or, } c = 84 – 52$
$\text{or, } c = 32\text{ cm}$
Semi-perimeter $s = \frac{84}{2} = 42\text{ cm}$.
$\text{Area of the triangle} = \sqrt{s(s – a)(s – b)(s – c)}$
$= \sqrt{42(42 – 25)(42 – 27)(42 – 32)}$
$= \sqrt{42 \times 17 \times 15 \times 10}$
$= \sqrt{107100}$
$\approx 327.26$
Answer:$327.26\text{ sq. cm}$ (approx.).
5. When the length of each side of an equilateral triangle is increased by $2\text{ metre}$, its area increases by $6\sqrt{3}\text{ square metre}$. Find the length of the side of the triangle.
Solution: Let, the length of each side of the equilateral triangle be $a\text{ m}$.
Therefore, the area of the equilateral triangle $= \frac{\sqrt{3}}{4}a^2\text{ sq. m}$.
If the side length is increased by $2\text{ m}$, the new side length becomes $(a + 2)\text{ m}$.
Therefore, the new area $= \frac{\sqrt{3}}{4}(a + 2)^2\text{ sq. m}$.
According to the question,
$\frac{\sqrt{3}}{4}(a + 2)^2 – \frac{\sqrt{3}}{4}a^2 = 6\sqrt{3}$
$\text{or, } \frac{\sqrt{3}}{4} \left\{(a + 2)^2 – a^2\right\} = 6\sqrt{3}$
$\text{or, } \frac{1}{4} (a^2 + 4a + 4 – a^2) = 6 \quad \text{[Dividing both sides by } \sqrt{3}\text{]}$
$\text{or, } \frac{1}{4} (4a + 4) = 6$
$\text{or, } 4a + 4 = 24$
$\text{or, } 4a = 24 – 4$
$\text{or, } 4a = 20$
$\text{or, } a = \frac{20}{4}$
$\text{or, } a = 5$
Answer: The side length of the triangle is $5\text{ m}$.
6. The lengths of the two sides of a triangle are $26\text{ m}$ and $28\text{ m}$ respectively, and its area is $182\text{ sq. m}$. Find the angle between the two sides.
Solution: Let,
Length of the first side, $a = 26\text{ m}$
Length of the second side, $b = 28\text{ m}$
Included angle between the sides $= \theta$
We know that the area of a triangle $= \frac{1}{2} ab \sin\theta$
According to the question,
$\frac{1}{2} ab \sin\theta = 182$
$\text{or, } \frac{1}{2} \times 26 \times 28 \times \sin\theta = 182$
$\text{or, } 13 \times 28 \times \sin\theta = 182$
$\text{or, } 364 \sin\theta = 182$
$\text{or, } \sin\theta = \frac{182}{364}$
$\text{or, } \sin\theta = \frac{1}{2}$
$\text{or, } \sin\theta = \sin 30^\circ$
$\text{or, } \theta = 30^\circ$
Answer: The included angle between the sides is $30^\circ$.
7. The length of equal sides of an isosceles triangle is $10\text{ m}$ and area $48\text{ square metre}$. Find the length of the base.
Solution: Let, the length of each equal side be $a = 10\text{ m}$.
Length of the base $= b\text{ m}$.
We know that the area of an isosceles triangle $= \frac{b}{4} \sqrt{4a^2 – b^2}$
According to the question,
$\frac{b}{4} \sqrt{4(10)^2 – b^2} = 48$
$\text{or, } b \sqrt{400 – b^2} = 48 \times 4$
$\text{or, } b \sqrt{400 – b^2} = 192$
$\text{or, } b^2 (400 – b^2) = 192^2 \quad \text{[Squaring both sides]}$
$\text{or, } 400b^2 – b^4 = 36864$
$\text{or, } b^4 – 400b^2 + 36864 = 0$
$\text{or, } b^4 – 256b^2 – 144b^2 + 36864 = 0$
$\text{or, } b^2(b^2 – 256) – 144(b^2 – 256) = 0$
$\text{or, } (b^2 – 256)(b^2 – 144) = 0$
Either,
$b^2 – 256 = 0 \implies b^2 = 256 \implies b = \sqrt{256} = 16$
Or,
$b^2 – 144 = 0 \implies b^2 = 144 \implies b = \sqrt{144} = 12$
Answer: Base length is $12\text{ m}$ or $16\text{ m}$.
8. Two roads run from a certain place with an angle of $135^\circ$ in two directions. Two persons move from that place in two directions with the speeds of $7\text{ km}$ per hour and $5\text{ km}$ per hour respectively. What will be the direct distance between them after $4\text{ hours}$.
Solution: Let, the starting point be $O$.
After $4\text{ hours}$, the first person reaches point $A$ and the second person reaches point $B$.
Distance covered by the first person, $OA = 7 \times 4 = 28\text{ km}$.
Distance covered by the second person, $OB = 5 \times 4 = 20\text{ km}$.
Angle between the paths, $\angle AOB = 135^\circ$.
Applying the Cosine Rule to find the direct distance $AB$ after $4\text{ hours}$:
$AB^2 = OA^2 + OB^2 – 2 \cdot OA \cdot OB \cdot \cos(\angle AOB)$
$\text{or, } AB^2 = 28^2 + 20^2 – 2 \times 28 \times 20 \times \cos 135^\circ$
$\text{or, } AB^2 = 784 + 400 – 1120 \times \left(-\frac{1}{\sqrt{2}}\right)$
$\text{or, } AB^2 = 1184 + \frac{1120}{\sqrt{2}}$
$\text{or, } AB^2 = 1184 + 791.96$
$\text{or, } AB^2 = 1975.96$
$\text{or, } AB = \sqrt{1975.96} \approx 44.45$
Answer: The direct distance is approximately $44.45\text{ km}$.
9. If the length pf the perpendicular from an point interior of an equilateral triangle to three sides are $6\text{ cm}$, $7\text{ cm}$, and $8\text{ cm}$ respectively; Find the length of sides of the triangles and the area of the triangular region.
Solution: Let, the side length of the equilateral triangle $= a\text{ cm}$.
The perpendiculars drawn from the interior point $O$ to the sides are $p_1 = 6\text{ cm}$, $p_2 = 7\text{ cm}$, and $p_3 = 8\text{ cm}$.
We know that the altitude of an equilateral triangle, $h = p_1 + p_2 + p_3$
$\text{or, } \frac{\sqrt{3}}{2} a = 6 + 7 + 8$
$\text{or, } \frac{\sqrt{3}}{2} a = 21$
$\text{or, } a = \frac{21 \times 2}{\sqrt{3}}$
$\text{or, } a = \frac{42}{\sqrt{3}} = 14\sqrt{3} \approx 24.25\text{ cm}$
$\text{Area of the triangle} = \frac{\sqrt{3}}{4} a^2$
$= \frac{\sqrt{3}}{4} (14\sqrt{3})^2$
$= \frac{\sqrt{3}}{4} (196 \times 3)$
$= 147\sqrt{3} \approx 254.61\text{ sq. cm}$
Answer: Side length is $24.25\text{ cm}$ (approx.) and area is $254.61\text{ sq. cm}$ (approx.).
10. The perpendicular of a right-angled triangle is $6\text{ cm}$ less than $\frac{11}{12}$ times of the base, and the hypotenuse is $3\text{ cm}$ less than $\frac{4}{3}$ times of the base.
(1) Let the base be $x$. Express the area of the triangle in terms of $x$.
Solution: Given, Base $= x\text{ cm}$.
Perpendicular, $h = \left(\frac{11x}{12} – 6\right)\text{ cm}$.
We know that the area of a right-angled triangle $= \frac{1}{2} \times \text{Base} \times \text{Perpendicular}$
$= \frac{1}{2} \cdot x \cdot \left(\frac{11x}{12} – 6\right)$
$= \frac{x}{2} \left(\frac{11x – 72}{12}\right)$
$= \frac{11x^2 – 72x}{24}$
Answer:$\frac{11x^2 – 72x}{24}\text{ sq. cm}$.
(2) Find the length of the base.
Solution: Base $= x$
Perpendicular $= \frac{11x}{12} – 6$
Hypotenuse $= \frac{4x}{3} – 3$
According to Pythagoras’ theorem:
$\text{Base}^2 + \text{Perpendicular}^2 = \text{Hypotenuse}^2$
$\text{or, } x^2 + \left(\frac{11x}{12} – 6\right)^2 = \left(\frac{4x}{3} – 3\right)^2$
$\text{or, } x^2 + \left(\frac{11x – 72}{12}\right)^2 = \left(\frac{4x – 9}{3}\right)^2$
$\text{or, } x^2 + \frac{(11x – 72)^2}{144} = \frac{(4x – 9)^2}{9}$
$\text{or, } 144x^2 + (11x – 72)^2 = 16(4x – 9)^2 \quad \text{[Multiplying both sides by } 144\text{]}$
$\text{or, } 144x^2 + 121x^2 – 1584x + 5184 = 16(16x^2 – 72x + 81)$
$\text{or, } 265x^2 – 1584x + 5184 = 256x^2 – 1152x + 1296$
$\text{or, } 265x^2 – 256x^2 – 1584x + 1152x + 5184 – 1296 = 0$
$\text{or, } 9x^2 – 432x + 3888 = 0$$
$\text{or, } x^2 – 48x + 432 = 0 \quad \text{[Dividing both sides by } 9\text{]}$
$\text{or, } x^2 – 36x – 12x + 432 = 0$
$\text{or, } x(x – 36) – 12(x – 36) = 0$
$\text{or, } (x – 36)(x – 12) = 0$
Either,
$(x – 36) = 0 \implies x = 36$
Or,
$(x – 12) = 0 \implies x = 12$
If $x = 12$, Perpendicular $= \frac{11(12)}{12} – 6 = 5\text{ cm}$ and Hypotenuse $= \frac{4(12)}{3} – 3 = 13\text{ cm}$.
Alternatively, $x = 36$ is also an acceptable value.
Answer: Base length is $12\text{ cm}$ or $36\text{ cm}$.
(c) If the length of the base of the triangle is $12\text{ cm}$, find the area of an equilateral triangle having the same perimeter as its perimeters.
Solution: If Base, $x = 12\text{ cm}$:
Perpendicular $= \frac{11(12)}{12} – 6 = 11 – 6 = 5\text{ cm}$
Hypotenuse $= \frac{4(12)}{3} – 3 = 16 – 3 = 13\text{ cm}$
Perimeter of the right-angled triangle $= 12 + 5 + 13 = 30\text{ cm}$.
According to the question,
Perimeter of the equilateral triangle $= 30\text{ cm}$.
Let each side of the equilateral triangle $= a\text{ cm}$.
Therefore,
$3a = 30 \implies a = 10\text{ cm}$
$\text{Area of the equilateral triangle} = \frac{\sqrt{3}}{4} a^2$
$= \frac{\sqrt{3}}{4} (10)^2$
$= \frac{\sqrt{3}}{4} \times 100$
$= 25\sqrt{3} \approx 43.30$
Answer:$25\sqrt{3}\text{ sq. cm}$ or $43.30\text{ sq. cm}$ (approx.).