Case study (4 Marks)

Question 14 Marks

Engineers often use the familiar triangular shape for strength in bridge design. Triangles are effective tools for architecture and are used in the design of bridges, buildings and other structures as they provide strength and stability. The triangle is common in all sorts of building supports and trusses. Following are some questions on triangles:

(i) In triangles ABC and DEF, if AB = DE, AC = EF and $\angle A=\angle E$. Then,
(a) $\triangle A B C \cong \triangle D E F$ by SAS criterion $\quad$(b) $\triangle A B C \cong \triangle E F D$ by SSS criterion
(c) $\triangle A B C \cong \triangle E D F$ by SAS criterion $\quad$(d) $\triangle A B C \cong \triangle E D F$ by ASA criterion
(ii) If $\triangle P R Q \cong \triangle D E F$, then $D E=$
(a) PR $\quad$(b) RQ $\quad$(c) PQ $\quad$(d) DF
(iii) Is it possible to construct a triangle with lengths of sides as 5 cm, 4 cm and 10 cm ?
(iv) In triangles ABC and DEF, AB = FD and $\angle A=\angle D$. Then the two triangles will be congruent by SAS axiom, if
(a) BC = EF $\quad$(b) AC = DE $\quad$(c) AC = EF $\quad$(d) BC = DE

Answer

(i) (c): In $\triangle^{\prime} s A B C$ and DEF, we have
$A B=D E, \angle A=\angle F \text { and } A C=E F$
So, by $S A S$ congruence criterion $\triangle A B C \cong \triangle E D F$.
(ii) (a): $\triangle P R Q \cong \triangle D E F \Rightarrow P R=D E$
(iii) No, sum of any two sides must be greater than the third side.
(iv) (b): $\triangle A B C \cong \triangle F D E$ by SAS criterion
$\therefore \quad A C=D E$

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Question 24 Marks

A ladder manufacturing company manufactures foldable step ladders of aluminum as shown in Fig. The lengths of two legs AB and AC are both equal to 110 cm and the angle between the two legs is $30^{\circ}$. On the basis of the above information answer the following questions:

(i) $\angle A B C$ is equal to
(a) $70^{\circ}$ $\quad$(b) $75^{\circ}$(c) $85^{\circ}$ $\quad$(d) $60^{\circ}$
(ii) If $\angle B A C=60^{\circ}$, then $B C=$
(a) 120 cm $\quad$(b) 55 cm $\quad$(c) 110 cm $\quad$(d) 100 cm
(iii) $\triangle A B C$ is
(a) isosceles acute angled $\quad$ (b) right angled isosceles
(c) isosceles obtuse angled$\quad$(d) equilateral
(iv) In two triangles ABC and DEF, if $\angle A=\angle D, A B=D E$ and $A C=D F$, then the criterion by which two triangles are congruent is
(a) SSS $\quad$(b) ASA $\quad$(c) AAS $\quad$(d) SAS

Answer

(i) (b): We have, AB = AC. So, $\triangle A B C$ is isosceles.
$\begin{array}{ll}\therefore & \angle A B C=\angle A C B \\
\text { Now, } & \angle B A C+\angle A B C+\angle A C B=180^{\circ} \Rightarrow 30^{\circ}+2\angle A B C=180^{\circ} \Rightarrow \angle A B C=75^{\circ}\end{array}$
(ii) (c): If $\angle B A C=60^{\circ}$, then
$A B=A C \Rightarrow \angle A C B=\angle A B C$
$\begin{array}{ll}\therefore & \angle A B C+\angle A C B+\angle B A C=180^{\circ} \Rightarrow \angle A B C+\angle A B C+60^{\circ}=180^{\circ} \Rightarrow 2 \angle A B C=120^{\circ} \\
\Rightarrow & \angle A B C=60^{\circ}\end{array}$
Thus, we have, $\angle A B C=\angle A C B=\angle B A C=60^{\circ}$. So, $\triangle A B C$ is equilateral.
Hence, $A B=B C=A C \Rightarrow B C=110 cm$
(iii) (a): Since AB = AC, Therefore $\triangle A B C$ is isosceles acute angled triangle.
(iv) (d): We have, $\angle A=\angle D$ AB = DE and AC = DF. So, by SAS criterion of congruence, $\triangle A B C \cong \triangle D E F$.

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Maths Case study (4 Marks)

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