DAV Class 7 Maths Chapter 8 Worksheet 4
Triangle And Its Properties Worksheet 4
1. The hypotenuse of a right triangle is 17 cm long. If one of the remaining two sides is of length 8 cm, find the length of the third side.
Solution
\begin{align*} \text{In } & \triangle ABC \\ \text{(hypotenuse )} AC &= 17\text{ cm} \\ BC &= 8\text{ cm} \\ AB &= x \\ \\ \text{Using Pythagoras}& \text{ theorem} \\ (AB)^2 + (BC)^2 &= (AC)^2 \\ x^2 + 8^2 & = 17^2 \\ x^2 + 64 &= 289 \\ x^2 & = 289 - 64 \\ x^2 & = 225 \\ x^2 & = 15^2 \\ x &= 15 \\ \color{green} AB &= \color{green} 15 \, cm \end{align*}
Answer Length of the third side \( = \color{red}15 \text{ cm} \)
2. The length of the hypotenuse of a right triangle is 13 cm. If one of the side of the triangle be 5 cm long, find the length of the other side.
Solution
\begin{align*} \text{In } \triangle & ABC, \text{ Let} \\ \text{(hypotenuse )} AC &= 13\text{ cm} \\ BC &= 5\text{ cm} \\ AB &= x\text{ cm} \\ \\ \text{Using Pythagoras theorem} \\ (AB)^2 + (BC)^2 & = (AC)^2\\ x^2 + 5^2 & = 13^2 \\ x^2 + 25 &= 169 \\ x^2 &= 169 - 25\\ x^2 &= 144 \\ x^2 &= 12^2 \\ x &= 12 \\ \color{green} AB &= \color{green} 12 \text{ cm} \end{align*}
Answer Length of the other side \( = \color{red}13 \text{ cm} \)
3. The length of two sides of a right triangle are equal. The square of the hypotenuse is \( \color{black} 800 \, cm^2 \). Find the length of the each side.
Solution
\begin{align*} \text{In } \triangle & ABC, \text{ Let} \\ AB &= BC \\ AC^2 &= 800 \text{ cm}^2 \\ AB &= x \\ BC &= x \\ \\ \text{Using Pythagoras theorem} \\ (AB)^2 + (BC)^2 &= (AC)^2 \\ x^2 + x^2 &= 800 \\ 2x^2 &= 800 \\ x^2 &= \frac{800}{2} \\ x^2 &= 400 \\ x &= 20^2 \\ x &= 20 \, \text{cm} \\ \implies \color{green} AB = 20 \, \text{cm} \, &, \, \color{green} BC = 20 \, \text{cm} \end{align*}
Answer Length of each side \( = \color{red}20 \text{ cm} \)
4. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
Solution
\begin{align*} \text{In } \triangle CDE, & \text{ Let} \\ DE &= 11 - 6 \\ &= 5m \\ EC &= 12 m \\ CD &= x \\ \\ &\text{Using Pythagoras theorem} \\ (CD)^2 &= (DE)^2 + (EC)^2 \\ x^2 &= 5^2 + (12)^2 \\ x^2 &= 25 + 144 \\ x^2 &= 169 \\ x^2 &= (13)^2 \\ x &= 13 \\ \color{green} CD &= \color{green} 13 m \\ \end{align*}
Answer Distance between their tops \( = \color{red} 13 m \)
5. Find the length of the diagonal of a rectangle whose sides are 15 cm and 8 cm.
Solution
\begin{align*} &\text{In right angle triangle DAB} \\ AD & = 8 \, cm \\ AB & = 15 \, cm \\ DB & = x \\ &\text{Using Pythagoras theorem} \\ (DB)^2 &= (AB)^2 + (AD)^2 \\ x^2 &= (15)^2 + (8)^2 \\ x^2 &= 225 + 64 \\ x^2 &= 289 \\ x^2 &= (17)^2 \\ x &= 17 \\ \color{green} DB &= \color{green} 17 \, cm \\ \end{align*}
Answer Length of the diagonal \( = \color{red} 17 cm \)
6. The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its top reach?
Solution
\begin{align*} \text{In } \triangle ABC \\ \text{Using } &\text{Pythagoras theorem} \\ (AC)^2 &= (AB)^2 + (BC)^2 \\ x^2 &= (6)^2 + (8)^2 \\ x^2 &= 36 + 64 \\ x^2 &= 100 \\ x^2 &= (10)^2 \\ x &= 10 \\ \color{green} AC &= \color{green} 10 \, m \\ \end{align*}
\begin{align*} \text{In } \triangle DEF \\ \text{Using Pythagoras} &\text{ theorem} \\ (FE)^2 &= (DE)^2 + (DF)^2 \\ (10)^2 &= (8)^2 + x^2 \\ 100 &= 64 + x^2 \\ 100 - 64 &= x^2 \\ 36 &= x^2 \\ (6)^2 &= x^2 \\ (6)&= x \\ \color{green} DF &= \color{green} 6 \, m \\ \end{align*}
Answer Height \( = \color{red} 6 m \)
7. A right angled triangle is isosceles. If the square of the hypotenuse is \({ 50 m^2 }\), what is the length of each of its sides?
Solution
\begin{align*} \text{In } & \triangle ABC \\ AB &= BC \\ AB &= x \\ BC &= x \\ (AC)^2 &= 50m^2 \\ \\ \text{Using Pythagoras} &\text{ theorem} \\ (AB)^2 + (BC)^2 & = (AC)^2 \\ x^2 + x^2 & = 50 \\ 2x^2 & = 50 \\ x^2 & = \frac{50}{2} \\ x^2 & = 25 \\ x^2 & = 5^2 \\ x & = 5 \\ AB &= 5m \\ BC &= 5m \\ \end{align*}
Answer Length of each side \( = \color{red} 5m \)
8. If the sides of a triangle are 3 m, 4 m and 6 m long, determine whether the triangle is a right angled triangle?
Solution
\begin{align*} \text{In } \triangle &ABC \text{, Let} \\ AC &= 6m \text{(Longest side)} \\ AB &= 4m \\ BC &= 3m \\ \text{According to the } &\text{Pythagoras theorem} \\ (AC)^2 &= (AB)^2 + (BC)^2 \\ \\ (AC)^2 &= (6)^2 \\ \color{green}(AC)^2 &= \color{green}36 \\ \\ (AB)^2 + (BC)^2 &= (4)^2 + (3)^2 \\ (AB)^2 + (BC)^2 &= 16 + 9 \\ \color{green}(AB)^2 + (BC)^2 &=\color{green} 25 \\ \\ 36 &\neq 25 \\ (AC)^2 &\neq (AB)^2 + (BC)^2 \\ \end{align*}
Answer The given triangle is \( \color{red} not \) a right angled triangle.
9. Verify whether the following triplets are Pythagorean or not.
(a) (8, 9, 10)
Solution
\begin{align*} \text{Let} \\ a &= 8 \\ b &= 9 \\ c &= 10 \\ \\ \text{According to the } &\text{Pythagorean triplet} \\ c^2 &= a^2 + b^2 \\ \\ \end{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline (10)^2 & (8)^2 + (9)^2 \\ = 100 & = 64 + 81 \\ & = 145 \\ \end{array} \begin{align*} 100 &\neq 145 \end{align*}
Answer (8, 9, 10) is \( \color{red} not \) a Pythagorean triplet.
(b) (5, 7, 12)
Solution
\begin{align*} \text{Let} \\ a &= 5 \\ b &= 7 \\ c &= 12 \\ \\ \text{According to the } &\text{Pythagorean triplet} \\ c^2 &= a^2 + b^2 \\ \\ \end{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline (12)^2 & (5)^2 + (7)^2 \\ = 144 & = 25 + 49 \\ & = 74 \\ \end{array} \begin{align*} 144 &\neq 74 \end{align*}
Answer (5, 7, 12) is \( \color{red} not \) a Pythagorean triplet.
(c) (3, 4, 5)
Solution
\begin{align*} \text{Let} \\ a &= 3 \\ b &= 4 \\ c &= 5 \\ \\ \text{According to the } &\text{Pythagorean triplet} \\ c^2 &= a^2 + b^2 \\ \\ \end{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline (5)^2 & (3)^2 + (4)^2 \\ = 25 & = 9 + 16 \\ & = 25 \\ \end{array} \begin{align*} 25 &= 25 \end{align*}
Answer \( {\color{red} \text{Yes }} (3, 4, 5) \) is a Pythagorean triplet.
(d) (5, 12, 13)
Solution
\begin{align*} \text{Let} \\ a &= 5 \\ b &= 12 \\ c &= 13 \\ \\ \text{According to the } &\text{Pythagorean triplet} \\ c^2 &= a^2 + b^2 \\ \\ \end{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline (13)^2 & (5)^2 + (12)^2 \\ = 169 & = 25 + 144 \\ & = 169 \\ \end{array} \begin{align*} 169 &= 169 \end{align*}
Answer \({\color{red} \text{Yes }} (5, 12, 13) \) is a Pythagorean triplet.
(e) (9, 12, 15)
Solution
\begin{align*} \text{Let} \\ a &= 9 \\ b &= 12 \\ c &= 15 \\ \\ \text{According to the } &\text{Pythagorean triplet} \\ c^2 &= a^2 + b^2 \\ \\ \end{align*} \begin{array}{l|l} \text{LHS} & \text{RHS} \\ \hline (15)^2 & (9)^2 + (12)^2 \\ = 225 & = 81 + 144 \\ & = 225 \\ \end{array} \begin{align*} 225 &= 225 \end{align*}
Answer \({\color{red} \text{Yes }} (9, 12, 15) \) is a Pythagorean triplet.