DAV Class 7 Maths Chapter 8 Enrichment Question
Triangle And Its Properties Enrichment Question
1. In the given figure, AB = AC and BC = CD, find x, y, z.
Solution
\begin{align*} In \ & \triangle ABC \\ \angle A & = 40^\circ \\ AB &= AC \\ \color{magenta} \text{Angles opposite to equal } & \color{magenta} \text{sides of a triangle are equal}\\ \angle ABC &= \angle ACB \\ \therefore \angle x &= \angle y \\ \\ \angle A + \angle x + \angle y & = 180^\circ \color{magenta} \text{ (Angle sum property)} \\ 40^\circ + \angle x + \angle x & = 180^\circ \\ 40^\circ + 2 \angle x & = 180^\circ \\ 2 \angle x & = 180^\circ - 40^\circ \\ 2 \angle x & = 140^\circ \\ \angle x & = \frac{140^\circ}{2} \\ \color{green} \angle x & = \color{green} 70^\circ \\ \color{green} \angle y & = \color{green} 70^\circ \\ \\ In \ & \triangle BCD \\ \angle BCD & = 90^\circ \\ CB &= CD \\ \color{magenta} \text{Angles opposite to equal } & \color{magenta} \text{sides of a triangle are equal}\\ \angle CBD &= \angle CDB \\ \\ \angle BCD + \angle CBD + \angle CDB & = 180^\circ \color{magenta} \text{ (Angle sum property)} \\ 90^\circ + \angle CBD + \angle z & = 180^\circ \\ 90^\circ + \angle z + \angle z & = 180^\circ \\ 90^\circ + 2 \angle z & = 180^\circ \\ 2 \angle z & = 180^\circ - 90^\circ \\ 2 \angle z & = 90^\circ \\ \angle z & = \frac{90^\circ}{2} \\ \color{green} \angle z & = \color{green} 45^\circ \end{align*}
Answer \( x = {\color{red} 70^\circ}, y = {\color{red} 70^\circ}, z = {\color{red} 45^\circ} \)
2. Two poles 15 m and 30 m high are standing straight on the ground. If their feet are 36 m apart, find the distance between their tops.
Solution
\begin{align*} \text{In } \triangle DEC, & \text{ Let} \\ DE &= 30 - 15 \\ &= 15m \\ EC &= 36 m \\ CD &= x \\ \\ &\text{Using Pythagoras theorem} \\ (CD)^2 &= (DE)^2 + (EC)^2 \\ x^2 &= (15)^2 + (36)^2 \\ x^2 &= 225 + 1296 \\ x^2 &= 1521 \\ x^2 &= (39)^2 \\ x &= 39 \\ \color{green} CD &= \color{green} 39 m \\ \end{align*}
Answer Distance between their tops \( = \color{red} 39 m \)