Question

Construct a triangle $ABC$ in which $BC=7cm,\angle B={75}^{\circ }$ and $AB+AC=13cm$

Answer 1

Given:

In $△ABC,BC=7$cm,$\angle B={75}^{\circ }$ and $AB+AC=13$cm

Method 1:

Steps of construction:

$1.$Draw the base $BC=7$cm

$2.$At the point $B$ first make an angle of ${90}^{\circ }$ and then bisect the angle between ${90}^{\circ }$ and ${60}^{\circ }$ to get $\angle XBC={75}^{\circ }$ (As shown in the figure)

$3.$Cut a line segment $BD$ equal to $AB+AC=13$cm from the ray $BX$

$4.$Join $DC$

$5.$ Now, draw a perpendicular bisector of line $CD$ with the help of a compass.

$6.$ Let that perpendicular bisector intersect line $BD$ at point $A$

$7.$ Join $AC$.

Thus, $△ABC$ has been constructed.

Method 2:

Steps of construction:

i) Draw a base $BC=7cm$.

ii) Make $\angle CBX$ measuring ${75}^{\circ }$ using a protractor.

iii) With $B$ as center and radius equal to $13cm$, draw an arc to intersect ray $BX$ at $D$.

iv) Join $DC$.

v) Measure $\angle D$ and make $\angle ACD=\angle D$. [Since sides opposite to equal angles are equal, $AC=AD$.]

vi) Let $CY$ intersect $BD$ at $A$.

$ABC$ is the required triangle.

Verification:

We see that,

$BC=7cm,\angle B={75}^{\circ }$ and $AB+AC=13cm$.