Question 14 Marks
The figure below consists of a square and an equilateral triangle connected together with a
common side.

In the design, $DF$ and $IG$ are two iron rods perpendicular to $BC.$ The measure of $\angle BAC = 120^\circ .$
$8.$ Which type of triangle is $ABC?$ Why$?$
$9.$ The central triangle AFG is equilateral. What is the measure of $\angle FDA?$
$A. 30^\circ $
$B. 60^\circ $
$C. 90^\circ $
$D. 120^\circ $
$10.$ The length of $IG$ is half of the length of $GC$. Write a proof for the statement.
common side.

In the design, $DF$ and $IG$ are two iron rods perpendicular to $BC.$ The measure of $\angle BAC = 120^\circ .$
$8.$ Which type of triangle is $ABC?$ Why$?$
$9.$ The central triangle AFG is equilateral. What is the measure of $\angle FDA?$
$A. 30^\circ $
$B. 60^\circ $
$C. 90^\circ $
$D. 120^\circ $
$10.$ The length of $IG$ is half of the length of $GC$. Write a proof for the statement.
Answer
View full question & answer→$8.$ Writes either isosceles or obtuse or both. Reasoning involves symmetry or measure of angle or both.
● Isosceles, as the design is symmetrical.
● Obtuse, as one of the angle is greater than $90^\circ .$
$9. D. 120^\circ $
$10.$ Valid mathematical proof involving properties of triangles.
● $IG$ is perpendicular to $BC,$ thus triangle $IGC$ is a right-angled triangle.
Measure of $\angle ICG = 30^\circ .$
Hence, $\angle CIG = 60^\circ .$
The sides of the triangle $IGC$ are in the ratio $2:1.$
● Isosceles, as the design is symmetrical.
● Obtuse, as one of the angle is greater than $90^\circ .$
$9. D. 120^\circ $
$10.$ Valid mathematical proof involving properties of triangles.
● $IG$ is perpendicular to $BC,$ thus triangle $IGC$ is a right-angled triangle.
Measure of $\angle ICG = 30^\circ .$
Hence, $\angle CIG = 60^\circ .$
The sides of the triangle $IGC$ are in the ratio $2:1.$










