Construct a triangle shadow similar to the given △ A B C , with its sides equal to 5 3 of the corresponding sides of the triangle A B C .

Given, △ A B C and the scale factor is 5 3 . So, scale factor > 1 . Steps of construction: 1 . Draw any ray AX making an acute angle with BC on the side opposite to the vertex A . 2 . Mark five points B 1 , B 2 , B 3 , B 4 , B 5 on AX such that BB 1 = B 1 B 2 = B 2 B 3 = B 3 B 4 = B 4 B 5 . 3 . Join B 3 C and draw a line through B 5 parallel to B 3 C to intersect BC extended at C ' . 4 . Draw a line through C ' parallel to the line AC to intersect BA extended at A ' . Therefore, △ AB ' C ' is the required triangle.

Construct a triangle shadow similar to the given △ A B C , with its sides equal to 5 3 of the corresponding sides of the triangle A B C .

Given, △ A B C and the scale factor is 5 3 . So, scale factor > 1 . Steps of construction: 1 . Draw any ray AX making an acute angle with BC on the side opposite to the vertex A . 2 . Mark five points B 1 , B 2 , B 3 , B 4 , B 5 on AX such that BB 1 = B 1 B 2 = B 2 B 3 = B 3 B 4 = B 4 B 5 . 3 . Join B 3 C and draw a line through B 5 parallel to B 3 C to intersect BC extended at C ' . 4 . Draw a line through C ' parallel to the line AC to intersect BA extended at A ' . Therefore, △ AB ' C ' is the required triangle.

Using pair of compasses and ruler, construct Δ PQR with sides PQ = 6 . 5 cm , QR = 7 cm and PR = 6 cm . Then construct Δ P ' QR ' whose sides are 4 3 of the corresponding sides of Δ PQR .

Given, Using pair of compasses and ruler, construct Δ PQR with sides PQ = 6 . 5 cm , QR = 7 cm and PR = 6 cm . Then construct Δ P ' QR ' whose sides are 4 3 of the corresponding sides of Δ PQR . Now, in order to construct Δ PQR , we follow the following steps: First, draw the line segment Q R of length 7 c m Taking P R as a radius, and keeping the compass at point R , we draw an arc of length 6 c m . Taking P Q as a radius, and keeping the compass at point Q , we draw an arc of length 6 . 5 cm. Then, the arcs P Q , P R intersects at point P . Join the points to obtain P Q , and P R . Thus, △ P Q R is constructed. Now, we need to construct Δ P ' QR ' whose sides are 4 3 of the corresponding sides of Δ PQR . Steps for construction: Draw any ray Q X making an acute angle with Q R on the side opposite to the vertex P . Locate 4 points (the greater of 4 , and 3 is 4 ) B 1 , B 2 , B 3 , and B 4 on Q X so that Q B 1 = B 1 B 2 = B 2 B 3 = B 3 B 4 Join B 3 to R and draw a line through B 4 parallel to B 3 R , intersecting the extended line segment Q R at R ' . Draw a line through R ' parallel to P R intersecting the extended line segment P Q at P ' . Thus, P ' Q R ' is the required triangle.

Using a pair of compasses and ruler, construct ∆ ABC with sides AB = 8 cm , BC = 9 . 5 cm , and A C = 6 . 5 cm . Then construct ∆ A ' BC ' whose sides are 3 4 of the corresponding sides of ∆ ABC .

Given, Using pair of compasses and ruler, construct Δ A B C with sides A B = 8 cm , B C = 9 . 5 cm and A C = 6 . 5 cm . Then construct ∆ A ' BC ' whose sides are 3 4 of the corresponding sides of Δ A B C . Now, in order to construct Δ A B C , we follow the following steps: First, draw the line segment B C of length 9 . 5 c m Taking A C as a radius, and keeping the compass at point C , we draw an arc of length 6 . 5 c m . Taking A B as a radius, and keeping the compass at point A , we draw an arc of length 8 cm. Then, the arcs of lengths A B , A C intersects at point A . Join the points to obtain A B , and A C . Thus, △ A B C is constructed. Now, we need to construct Δ A ' B C ' whose sides are 3 4 of the corresponding sides of Δ A B C . Steps for construction: Draw any ray B X making an acute angle with B C on the side opposite to the vertex A . Locate 4 points (the greater of 4 , and 3 is 4 ) B 1 , B 2 , B 3 , and B 4 on B X so that B B 1 = B 1 B 2 = B 2 B 3 = B 3 B 4 Join B 4 to C and draw a line through B 3 parallel to B 4 C , intersecting the line segment B C at C ' . Draw a line through C ' parallel to A C intersecting the line segment A B at A ' . Thus, A ' B C ' is the required triangle.

Draw a circle with centre O and radius 2 . 7 c m . Take a point P at a distance 8 . 5 c m from the centre of the circle. Using a pair of compasses and ruler, construct two tangents P S and P T to the circle. Measure and state the length of the tangent segments.

Given that we need to draw a circle whose centre is O and radius is 2 . 7 c m . Taking a point P at a distance 8 . 5 c m from the centre of the circle, we need to draw the tangents P S and P T from the point P to this circle using a pair of compasses and ruler. Steps of construction are as follows: First, draw a circle of radius 2 . 7 c m , whose centre is O Take a point P at a distance 8 . 5 c m from the centre of the circle, and join the line P O . Taking M as centre and M O as radius, draw a circle. Let it intersect the given circle at the points S and T . Join P S and P T Then P S and P T are the required two tangents. Since the tangent at any point are perpendicular to the radius. So, we have O T ⊥ P T , and O S ⊥ P S Now, in right triangle O T P , by Pythagoras Theorem, we have O P 2 = O T 2 + P T 2 ⇒ 8 . 5 2 = 2 . 7 2 + P T 2 ⇒ 72 . 25 - 7 . 29 = P T 2 ⇒ P T 2 = 64 . 96 c m 2 Hence, P T = 8 . 06 c m We know that the tangents drawn from an external point to the circle are of equal length ⇒ P T = P S = 8 . 06 c m