DAV Class 7 Maths Chapter 9 Worksheet 4
Congruent Triangles Worksheet 4
1. \( \triangle ABC \) is isosceles with AB = AC. AD is the altitude from A on BC.
(i) Is \( \triangle ABD \cong \triangle ACD \) ? Why ?
Answer
\[ \begin{align*} \color{magenta} Given : & \triangle ABC \text{ is an isosceles triangle}\\ & \therefore AB = AC \\ & AD \text{ is an altitude from A on BC} \\ \\ \color{magenta} \text{To prove} : & \triangle ABD \cong \triangle ACD \\ \\\text{In } \triangle ABD & \text{ and } \triangle ACD \\ \angle \text{ADB} &= \angle \text{ADC} \implies 90^\circ \ \ \color{green} R \\ \text{AB} &= \text{AC} \implies \text{(Given) }\color{green} H\\ \text{AD} &= \text{AD} \implies \text{(Common side) }\color{green} S\\ \text{By RHS} & \text{ congruence condition,} \\ & \triangle ABD \cong \triangle ACD \end{align*} \]
(ii) State the three pairs of matching parts you have used to answer (i)
\[ \begin{align*} \angle \text{ADB} &= \angle \text{ADC} \implies 90^\circ \ \ \color{green} R \\ \text{AB} &= \text{AC} \implies \text{(Given) }\color{green} H\\ \text{AD} &= \text{AD} \implies \text{(Common side) }\color{green} S\\ \end{align*} \]
(iii) Is it true to say that BD = CD? Why?
Answer Yes BD = CD, by corresponding parts of congruent triangles (C.P.C.T).
2. Which pairs of triangles are congruent by RHS congruence condition in the given figure? If congruent, write the congruence of the two triangles in symbolic form.
(i)
Solution
\[ \begin{align*} \text{In } \triangle ABD & \text{ and } \triangle BAC \\ \angle \text{ADB} &= \angle \text{BCA} \implies 90^\circ \ \ \color{green} R \\ \text{AB} &= \text{BA} \implies \text{(Common side) }\color{green} H\\ \text{AD} &= \text{BC} \implies 3 \,cm \text{ (Given) }\color{green} S\\ \text{By RHS} & \text{ congruence condition,} \\ & \triangle ABD \cong \triangle BAC \end{align*} \]
Answer \( \color{red} \triangle ABD \cong \triangle BAC \)
(ii)
Solution
\[ \begin{align*} \text{In } \triangle ABD & \text{ and } \triangle CBD \\ \angle \text{ADB} &= \angle \text{CDB} \implies 90^\circ \ \ \color{green} R \\ \text{AB} &= \text{CB} \implies \text{(Given) }\color{green} H\\ \text{BD} &= \text{BD} \implies \text{(Common side) }\color{green} S\\ \text{By RHS} & \text{ congruence condition,} \\ & \triangle ABD \cong \triangle CBD \end{align*} \]
Answer \( \color{red} \triangle ABD \cong \triangle CBD \)
(iii)
Solution
\[ \begin{align*} \text{In } \triangle ABD & \text{ and } \triangle ACD \\ \angle \text{ADB} &= \angle \text{ADC} \implies 90^\circ \ \ \color{green} R \\ \text{AB} &= \text{AC} \implies 4.5 \,cm \text{ (Given) }\color{green} H\\ \text{AD} &= \text{AD} \implies \text{(Common side) }\color{green} S\\ \text{By RHS} & \text{ congruence condition,} \\ & \triangle ABD \cong \triangle ACD \end{align*} \]
Answer \( \color{red} \triangle ABD \cong \triangle ACD \)
(iv)
Solution
\[ \begin{align*} \text{In } \triangle ABC & \text{ and } \triangle ADC \\ \angle \text{ABC} &= \angle \text{ADC} \implies 90^\circ \ \ \color{green} R \\ \text{AC} &= \text{AC} \implies \text{(Common side) }\color{green} H\\ \text{BC} &= \text{DC} \implies 6.4 \ cm \ \text{(Given) }\color{green} S\\ \text{By RHS} & \text{ congruence condition,} \\ & \triangle ABC \cong \triangle ADC \end{align*} \]
Answer \( \color{red} \triangle ABC \cong \triangle ADC \)
(v)
Solution
\[ \begin{align*} \text{In } \triangle AOB & \text{ and } \triangle COD \\ \angle \text{ABO} &= \angle \text{CDO} \implies 90^\circ \ \ \color{green} R \\ \text{AO} &= \text{CO} \implies 5 \ cm \ \text{(Given) }\color{green} H\\ \text{AB} &= \text{CD} \implies 3 \ cm \ \text{(Given) }\color{green} S\\ \text{By RHS} & \text{ congruence condition,} \\ & \triangle AOB \cong \triangle COD \end{align*} \]
Answer \( \color{red} \triangle AOB \cong \triangle COD \)
3. QS and RT are the altitudes of \( \triangle PQR \) and QS = RT as shown in the diagram.
(i) Is \( \triangle QRS \cong \triangle RQT \) by RHS congruence condition?
Answer
\[ \begin{align*} \color{magenta} Given : & \text{QS and RT are the altitudes of }\triangle PQR \\ & QS = RT \\ \\ \color{magenta} \text{To prove} : & \triangle QRS \cong \triangle RQT \\ \\\text{In } \triangle QRS & \text{ and } \triangle RQT \\ \angle \text{QSR} &= \angle \text{RTQ} \implies 90^\circ \ \ \color{green} R \\ \text{QR} &= \text{RQ} \implies \text{(Common side) }\color{green} H\\ \text{QS} &= \text{RT} \implies \text{(Given) }\color{green} S\\ \text{By RHS} & \text{ congruence condition,} \\ & \triangle QRS \cong \triangle RQT \end{align*} \]
(ii) State the three pairs of corresponding parts which make \( \triangle QRS \cong \triangle RQT \).
Answer
\[ \begin{align*} \text{In } \triangle QRS & \text{ and } \triangle RQT \\ \angle \text{QSR} &= \angle \text{RTQ} \implies 90^\circ \ \ \color{green} R \\ \text{QR} &= \text{RQ} \implies \text{(Common side) }\color{green} H\\ \text{QS} &= \text{RT} \implies \text{(Given) }\color{green} S\\ \end{align*} \]
4. In the given figure, \( \angle B = \angle P = 90^\circ \) and \( side \ AB = side \ PR \). What additional information is required to make \( \triangle ABC \cong \triangle RPQ \) by RHS congruence condition?
Answer
\[ \begin{align*} \text{In } \triangle ABC & \text{ and } \triangle RPQ \\ \angle \text{ABC} &= \angle \text{RPQ} \implies 90^\circ \ \ \color{green} R \\ \color{magenta} \text{AC} &= \color{magenta} \text{RQ} \implies \text{(Required side) } \color{green} H\\ \text{AB} &= \text{RP} \implies 4 \ cm \text{ (Given) }\color{green} S \\ \\ \end{align*} \]
AC = RQ is the additional information required to make \( \triangle ABC \cong \triangle RPQ \) by RHS congruence condition
5. In the given figure, \( AC = DB \) and \( AB \perp BC \) also \( DC \perp BC\). State which of the following statements is true. (i) \( \triangle ABC \cong \triangle DBC \) (ii) \( \triangle ABC \cong \triangle DCB \) (iii) \( \triangle ABC \cong \triangle BCD \) State the three pairs of matching parts you have used to arrive at the answer. Which side is equal to side AB? Why?
Answer
\[ \begin{align*} \text{In } \triangle ABC & \text{ and } \triangle DCB \\ \angle \text{ABC} &= \angle \text{DCB} \implies 90^\circ \ \ \color{green} R \\ \text{AC} &= \text{DB} \implies \text{(Given) } \color{green} H\\ \text{BC} &= \text{CB} \implies \text{(Common Side) }\color{green} S \\ \\ \therefore \triangle ABC & \cong \triangle DCB \ \boxed{(ii)} \text{ is true} \\ \\ AB = DC & \text{ by CPCT} \end{align*} \]
AC = RQ is the additional information required to make \( \triangle ABC \cong \triangle RPQ \) by RHS congruence condition