Question

Construct a triangle $ABC$, in which $BC=7cm,\angle A={70}^{\circ }$ and foot of the perpendicular $D$ on $BC$ from $A$ is $4.5cm$ away from $B$

Answer 1

Given that:

$BC=7cm.m\angle A={70}^{\circ },BD=4.5cm$

Steps of construction:

$1.$ Draw a line segment $BC=7cm$ and draw $\overline{BX}$ such that $m\angle CBX={70}^{\circ }$ and also draw $\overline{BY}$ such that $m\angle XBY={90}^{\circ }.$

$2.$ Draw the perpendicular bisector of $BC$ such that it intersects $\overline{BY}$ at ${}^{\prime }{O}^{\prime }.$

$3.$ Taking ${}^{\prime }{O}^{\prime }$ as centre and $OB+OC$ as radius, draw a circle.

$4.$ Taking ${}^{\prime }{B}^{\prime }$ as centre cut $\overline{BC}$ at $D$ with the radius of $4.5cm$ and also by using protractor draw a perpendicular through $D$ such that it intersect the circle at ${}^{\prime }{A}^{\prime }.$

$5.$ Join $AB$ and $AC.$

Thus, $△ABC$ is the required triangle.

By, alternate segment theorem $m\angle BAC=m\angle CBX={70}^{\circ }.$