DAV Class 7 Maths Chapter 8 Worksheet 2

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DAV Class 7 Maths Chapter 8 Worksheet 2

Triangle And Its Properties Worksheet 2


1. In \( \color{black} \triangle ABC , \angle A = 44^\circ \). If \( \color{black} AB = AC \), find \( { \color{black} \angle B} \text{ and } { \color{black} \angle C} \).

Solution

\begin{align*} AB &= AC \\ \text{Angles opposite to equal }& \text{sides of a triangle are equal}\\ \angle B &= \angle C\\ \text{Let } \angle B \text{ and } & \angle C \text{ be } x \\ \text{By the angle sum }& \text{property}\\ \angle A + \angle B + \angle C &= 180^\circ\\ 44^\circ + x + x &= 180^\circ\\ 44^\circ + 2x &= 180^\circ\\ 2x &= 180^\circ - 44^\circ\\ 2x &= 136^\circ\\ x &= \frac{136^\circ}{2}\\ x &= 68^\circ\\ \end{align*}

Answer \( \angle B = {\color{red} 68^\circ} , \angle C = {\color{red} 68^\circ} \)

2. \( \color{black} \triangle PQR \) is an isosceles triangle with \( \color{black} PQ = PR \). If \( \color{black} \angle R = 42^\circ \), find the measure of \( \color{black} \angle P \).

Solution

\begin{align*} PQ &= PR \\ \text{Angles opposite to equal }& \text{sides of a triangle are equal.}\\ \angle Q &= \angle R\\ \angle R = 42^\circ \, , \,& \angle Q = 42^\circ\\ \text{By the angle sum }& \text{property}\\ \angle P + \angle Q + \angle R &= 180^\circ\\ \angle P + 42^\circ + 42^\circ &= 180^\circ\\ \angle P + 84^\circ &= 180^\circ\\ \angle P &= 180^\circ - 84^\circ\\ \angle P &= 96^\circ\ \ \end{align*}

Answer \( \angle P = \color{red} 96^\circ \)

3. \( \color{black} \triangle ABC \) is an isosceles triangle with \( \color{black} AB = AC \). If \( \color{black} \angle B = 40^\circ \), what are the measures of \( { \color{black} \angle A} \text{ and } { \color{black} \angle C} \)?

Solution

\begin{align*} AB &= AC \\ \angle B &= 40^\circ\\ \text{Angles opposite to equal }& \text{sides of a triangle are equal}\\ \angle B &= \angle C\\ \angle C &= 40^\circ\\ \text{By the angle sum }& \text{property}\\ \angle A + \angle B + \angle C &= 180^\circ\\ \angle A + 40^\circ + 40^\circ &= 180^\circ\\ \angle A + 80^\circ &= 180^\circ\\ \angle A &= 180^\circ - 80^\circ\\ \angle A &= 100^\circ\\ \end{align*}

Answer \( \angle C = {\color{red} 40^\circ} , \angle A = {\color{red} 100^\circ} \)

4. The vertical angle of an isosceles triangle is \( \color{black} 100^\circ \). Find its base angles.

Solution

\begin{align*} \triangle ABC \text{ is an }& \text{Isosceles triangle} \\ AB &= AC \\ \angle A &= 100^\circ \\ \text{Angles opposite to equal }& \text{sides of a triangle are equal}\\ \angle B &= \angle C \\ \text{Let } \angle B \text{ and } & \angle C \text{ be } x \\ \text{By the angle sum }& \text{property}\\ \angle A + \angle B + \angle C &= 180^\circ \\ 100^\circ + x + x &= 180^\circ \\ 100^\circ + 2x &= 180^\circ \\ 2x &= 180^\circ - 100^\circ \\ 2x &= 80^\circ \\ x &= \frac{80^\circ}{2} \\ x &= 40^\circ \\ \end{align*}

Answer Base angles are \( \angle B = {\color{red} 40^\circ} , \angle C = {\color{red} 40^\circ} \)

5. In the diagram \( \color{black} \triangle ABC \) is isosceles with \( \color{black} AB=AC \) Find the values of \( \color{black} x \) and \( \color{black} y \).

Solution

\begin{align*} \triangle ABC \text{ is an }& \text{Isosceles triangle} \\ AB &= AC \\ \text{Angles opposite to equal }& \text{sides of a triangle are equal}\\ \angle ABC &= \angle ACB \\ \text{By the angle sum }& \text{property}\\ \angle A + \angle ABC + \angle ACB &= 180^\circ \\ 30^\circ + \angle ABC + \angle ABC &= 180^\circ \\ 30^\circ + 2\angle ABC &= 180^\circ \\ 2\angle ABC &= 180^\circ - 30^\circ \\ 2\angle ABC &= 150^\circ \\ \angle ABC &= \frac{150^\circ}{2} \\ \angle ABC &= 75^\circ \\ \\ \angle ABC = 75^\circ \, &, \, \angle ACB = 75^\circ \\ \\ \angle ABC + x^\circ &= 180^\circ \text{ (Linear Pair)}\\ 75^\circ + x^\circ &= 180^\circ \\ x^\circ &= 180^\circ - 75^\circ \\ x^\circ &= 105^\circ \\ \\ \angle ACB + y^\circ &= 180^\circ \text{ (Linear Pair)}\\ 75^\circ + y^\circ &= 180^\circ \\ y^\circ &= 180^\circ - 75^\circ \\ y^\circ &= 105^\circ \\\end{align*}

Answer \( x^\circ = {\color{red} 105^\circ} , y^\circ = {\color{red} 105^\circ} \)

6. In the given diagram, find \( \color{black} \angle ACB \) and \( \color{black} \angle A \) and write which sides of \( \color{black} \triangle ABC \) are equal.

Solution

\begin{align*} \angle ACB + \angle ACD &= 180^\circ \text{ (Linear Pair)}\\ \angle ACB + 130^\circ &= 180^\circ \\ \angle ACB &= 180^\circ - 130^\circ \\ \angle ACB &= 50^\circ \\ \\ \text{In }\triangle ABC \\ \text{By the angle sum }& \text{property}\\ \angle A + \angle B + \angle ACB &= 180^\circ\\ \angle A + 80^\circ + 50^\circ &= 180^\circ\\ \angle A + 130^\circ &= 180^\circ\\ \angle A &= 180^\circ - 130^\circ \\ \angle A &= 50^\circ \\ \angle A &= \angle ACB \\ BA &= BC \end{align*}

Answer The equal sides are \( \color{red} BA = BC\)

7. In the given diagram, \( \color{black} AB=AC \text{ and } AP \perp BC \text{ and } \angle B = 60^\circ \). Find (i) \( \color{black} \angle BAP \) (ii) \( \color{black} \angle ACB \)

Solution

\begin{align*} \text{In }\triangle ABP \\ \text{By the angle sum }& \text{property}\\ \angle BAP + \angle B + \angle APB &= 180^\circ\\ \angle BAP + 60^\circ + 90^\circ &= 180^\circ\\ \angle BAP + 150^\circ &= 180^\circ\\ \angle BAP &= 180^\circ - 150^\circ \\ \angle BAP &= 30^\circ \\ \\ \text{In }& \triangle ABC \\ AB &= AC \\ \text{Angles opposite to equal }& \text{sides of a triangle are equal}\\ \angle C &= \angle B \\ \angle ACB &= \angle ABC \\ \angle ACB &= 60^\circ\\ \end{align*}

Answer \( \angle BAP = {\color{red} 30^\circ} , \angle ACB = {\color{red} 60^\circ} \)

8. In the given diagram, find \( \color{black} x \) and \( \color{black} y \). Is \( \color{black} \triangle ABC \) isosceles? If so, name its equal sides.

Solution

\begin{align*} \angle ABX + \angle ABC &= 180^\circ \text{ (Linear Pair)}\\ 110^\circ + x^\circ &= 180^\circ \\ x^\circ &= 180^\circ - 110^\circ \\ x^\circ &= 70^\circ \\ \\\text{By the exterior }& \text{angle property}\\ \angle A + \angle C &= \angle ABX \\ 40^\circ + y^\circ &= 110^\circ\\ y^\circ &= 110^\circ - 40^\circ \\ y^\circ &= 70^\circ\\ \\ \text{In }\triangle ABC \\ \angle ABC &= \angle C \\ \text{Yes, }\triangle ABC & \text{ is isosceles.}\\ AB &= AC \\\end{align*}

Answer \( x^\circ = { \color{red} 70^\circ} , y^\circ = { \color{red} 70^\circ} \). \( \color{green} \text{Yes, }\triangle ABC \, \text{ is isosceles}. \) Equal sides are \( \color{red} AB= AC .\)