Question
Construct an isosceles triangle whose base is 9cm and altitude 5cm. Construct another triangle whose side are $\frac34$ of the corresponding sides of the first isosceles triangle.

Answer



Steps of construction:
  1. Draw a line segment BC = 9cm
  2. Draw peependicular bisector PQ of BC Meeting it at M.
  3. Fron QP cut-off a distance MA = 5cm.
  4. Join AB and AC.
Thus, isosceles $\triangle\text{ABC}$ is obtained.
  1. Below BC, make an acute $\angle\text{CBX}.$
  2. Along $B X$, mark off 4 points $B_1, B_2, B_3, B_4$, such that $B_1, B_1 B_2=B_1 B_3=B_3 B_4$.
  3. Join $\mathrm{B}_4 \mathrm{C}$
  4. From $B_3$, draw $B_3 C^{\prime} \| B_4 C$, meeting $B C$ at $C^{\prime}$.
  5. From C', draw C' A' || CA, meeting AB at A'.
Thus, $\triangle\text{A}'\text{BC}'$ is the required triangle similar to $\triangle\text{ABC}$ Such that each side of $\triangle\text{A}'\text{BC}'$ is $\frac34\text{times}$ the corresponding side of $\triangle\text{ABC}.$

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