DAV Class 7 Maths Chapter 9 Worksheet 2

Congruent Triangles Worksheet 2

1. In △ABC and △DEF, AB = DE and BC = EF as shown in the diagram. What additional information is required to make the two triangles congruent by SAS congruence condition?

Answer

In △ABC and △DEF.
\(\color{red} \angle B = \angle E \), is required to make the two triangles congruent by SAS congruence condition.

2. △ABC is isosceles with AB = AC as shown in the diagram. Line segment AD bisects \( \angle A \) and meets the base BC at D. Find the third pair of corresponding parts which make \( \triangle ADB \cong \triangle ADC \) by SAS congruence condition. Is it true to say that BD = CD? Why?

Answer

\[ \begin{align*} \text{In } \triangle ADB & \text{ and } \triangle ADC \\ \text{AB} &= \text{AC} \text{ (Given)} \color{red} (S)\\ \angle BAD &= \angle CAD \text{ ( AD bisects } \angle A \,) \color{red} (A)\\ \text{AD} &= \text{AD} \text{ (Common side)} \color{red} (S) \\ \\ \text{By SAS} & \text{ congruence condition,} \\ & \triangle ADB \cong \triangle ADC \\ \\ \text{Third pair } & \implies AD =AD \\ \text{Yes, } &BD = CD \text{ (by CPCT)} \end{align*} \]

3. In the below diagram AB \( \parallel \) DC and AB = DC.

(i) Is \( \angle \text{BAC} = \angle \text{DCA} \) ? Why ?

Answer \( \angle \text{BAC} = \angle \text{DCA}\) (Alternate angles)

(ii) Is \( \triangle \text{ABC} \cong \triangle \text{CDA} \) by SAS congruence condition?

Answer Yes, \( \triangle \text{ABC} \cong \triangle \text{CDA} \) by SAS congruence condition.

(iii) State the three facts you have used to answer (ii).

Answer

\[ \begin{align*} \text{In } \triangle ABC & \text{ and } \triangle CDA \\ \text{AB} &= \text{CD} \text{ (Given) } \color{red} (S) \\ \angle \text{BAC} &= \angle \text{DCA} \text{ (Alternate angles) } \color{red} (A) \\ \text{AC} &= \text{CA} \text{ (Common side) } \color{red} (S) \\ \\\text{By SAS} & \text{ congruence condition,} \\ & \triangle ABC \cong \triangle CDA \end{align*} \]

4. In the below diagrams, which pairs of triangles are congruent by SAS congruence condition? If congruent, write the congruence of the two triangles in symbolic form

(i)

Solution

\[ \begin{align*} \text{In } \triangle ABC & \text{ and } \triangle EDF \\ \text{AB} &= \text{ED} \implies (3 \,cm) \color{red} (S) \\ \angle \text{B} &= \angle \text{D} \implies (70^\circ ) \color{red} (A) \\ \text{BC} &= \text{DF} \implies (4 \,cm) \color{red} (S) \\ \\ \text{By SAS} & \text{ congruence condition,} \\ & \triangle ABC \cong \triangle EDF \end{align*} \]

Answer Yes, \( \color{red} \triangle \text{ABC} \cong \triangle \text{EDF} \)

(ii)

Solution

\[ \begin{align*} \text{In } \triangle PQR & \text{ and } \triangle ZXY \\ \text{PQ} &= \text{ZX} \implies (3 \,cm) \color{red} (S) \\ \angle \text{Q} &= \angle \text{X} \implies (90^\circ ) \color{red} (A) \\ \text{QR} &= \text{XY} \implies (2 \,cm) \color{red} (S) \\ \\ \text{By SAS} & \text{ congruence condition,} \\ & \triangle PQR \cong \triangle ZXY \end{align*} \]

Answer Yes, \( \color{red} \triangle PQR \cong \triangle ZXY \)

(iii)

Solution

\[ \begin{align*} \text{In } \triangle AOB & \text{ and } \triangle COD \\ \text{AO} &= \text{CO} \implies (Given) \color{red} (S) \\ \angle \text{AOB} &= \angle \text{COD} \implies \text{(Vertically opposite angles )} \color{red} (A) \\ \text{BO} &= \text{DO} \implies (Given) \color{red} (S) \\ \\ \text{By SAS} & \text{ congruence condition,} \\ & \triangle AOB \cong \triangle COD \end{align*} \]

Answer Yes, \( \color{red} \triangle \text{AOB} \cong \triangle \text{COD} \)

(iv)

Solution

\[ \begin{align*} \text{In } \triangle PQS & \text{ and } \triangle RSQ \\ \text{PS} &= \text{RQ} \implies (4 \, cm) \color{red} (S) \\ \angle \text{PSQ} &= \angle \text{RQS} \implies \text{(Alternate angles )} \color{red} (A) \\ \text{QS} &= \text{SQ} \implies (5.8 \, cm) \color{red} (S) \\ \\ \text{By SAS} & \text{ congruence condition,} \\ & \triangle PQS \cong \triangle RSQ \end{align*} \]

Answer Yes, \( \color{red} \triangle \text{PSQ} \cong \triangle \text{RQS} \)

(v)

Solution

\[ \begin{align*} \text{In } \triangle PRQ & \text{ and } \triangle XYZ \\ \text{PR} &= \text{XY} \implies (4 \,cm) \color{red} (S) \\ \angle \text{PRQ} &= \angle \text{XYZ} \implies (55^\circ ) \color{red} (A) \\ \text{RQ} &= \text{YZ} \implies (2 \,cm) \color{red} (S) \\ \\ \text{By SAS} & \text{ congruence condition,} \\ & \triangle PQR \cong \triangle XZY \end{align*} \]

Answer Yes, \( \color{red} \triangle PRQ \cong \triangle XYZ \)

(vi)

Solution

\[ \begin{align*} \text{In } \triangle ABD & \text{ and } \triangle BAC \\ \text{AD} &= \text{BC} \implies (Given) \color{red} (S) \\ \text{BD} &= \text{AC} \implies \text{(Given )} \color{red} (S) \\ \text{AB} &= \text{BA} \implies \text{(Common Side)} \color{red} (S) \\ \\ \text{By SSS} & \text{ congruence condition,} \\ & \triangle ABD \cong \triangle BAC \\ \\\text{In } \triangle ABD & \text{ and } \triangle BAC \\ \text{AD} &= \text{BC} \implies (Given) \color{red} (S) \\ \angle \text{D} &= \angle \text{C} \implies \text{(By CPCT)} \color{red} (A) \\ \text{BD} &= \text{AC} \implies (Given) \color{red} (S) \\ \\ \text{By SAS} & \text{ congruence condition,} \\ & \triangle ABD \cong \triangle BAC \\ \\ \end{align*} \]

Answer Yes, \( \color{red} \triangle ABD \cong \triangle BAC \)

5. In the given diagram, AB = AD and \( \angle BAC = \angle DAC \).

(i) State in symbolic form, the congruence of two triangles ABC and ADC is true. Also state the congruence condition used.

Answer

\[ \begin{align*} \text{In } \triangle ABC & \text{ and } \triangle ADC \\ \text{AB} &= \text{AD} \implies (Given) \color{red} (S) \\ \angle BAC &= \angle DAC \implies (Given) \color{red} (A) \\ \text{AC} &= \text{AC} \implies \text{(Common side)} \color{red} (S) \\ \\ \text{By SAS} & \text{ congruence condition,} \\ & \triangle ABD \cong \triangle BAC \end{align*} \]

(ii) Complete each of the following, so as to make it true:

(a) \( \angle ABC = \) \( \color{red} \angle ADC \)

(b) \( \angle ACD = \) \( \color{red} \angle ACB \)