Chapter 14 Construction Of A Triangle Exercise 14
Question 1. Pritam drew a quadrilateral ABCD of which AB = 5 cm, BC = 6 cm, CD = 4 cm, DA = 3 cm, and ∠ABC = 60°. I drew a triangle with an equal area of that quadrilateral.
Solution:
Construction: First, a straight line BX is drawn. From BX, BC is cut off equal to 6cm. On point B equal to a 60° angle ∠CBY is drawn. From BY equal to AB 5 is cut off Taking A as the center and taking a radius equal to 3 cm a radius is drawn. Again, taking C as a center and taking a radius equal to 4 cm, an arc is drawn.
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Both arcs intersect each other at the D point. A, D, and C, D are joined. Consequently, ABCD, a quadrilateral whose sides AB = 5 cm, BC = 6 cm, CD = 4 cm, AD = 3 cm, and ∠ABC 60°, is formed.
Diagonal AC is drawn D From point parallel to AC a st. line is drawn which cuts BX at point E. A, and E is joined. ABE is the required triangle whose area is equal to quadrilateral ABCD.
Question 2. Sahana drew a quadrilateral ABCD of which AB = 4 cm, BC = 5 cm, CD = 4.8 cm, DA = 4.2 cm and diagonal AC = 6 cm. I draw a triangle with an equal area of the quadrilateral.
Solution:
Equal in area to quadrilateral ABCD, triangle ABE is drawn.
Question 3. Sahana drew a rectangle ABCD, of which AB = 4 cm and BC = 6 cm. I draw a triangle with an equal area of that rectangle.
Solution:
Equal in the area to rectangle ABCD, triangle ADE is drawn.
Question 4. I draw a quadrilateral ABCD of which BC = 6 cm, AB = 4 cm, CD = 3 cm, ∠ABC 60°, ∠BCD = 55°. I draw a triangle with an equal area of that quadrilateral of which one side is alongside AB and another side is alongside BC.
Solution:
Equal in area to quadrilateral ABCD, triangle ABE is drawn.
Question 5. I draw a square with side of 5 cm. I draw a parallelogram of which one angle is 60°, equal area to the square.
Solution:
Equal in area to ABCD a square of side 5 cm ΔABE is drawn and equal in area to ΔABE a parallelogram CEGF is drawn whose angle ∠FCE = 60°.
Question 6. I draw a square with a side 6 cm and I draw a triangle with an equal area to that square.
Solution:
A square of side 6 cm, ABCD is drawn and equal in area to this square, a triangle ABE is drawn.
Question 7. I draw a quadrilateral ABCD, of which AD and BC are perpendicular on side AB and AB = 5 cm, AD = 7 cm, and BC = 4 cm. I draw a triangle with an equal area of that quadrilateral of which one angle is 30°.
Hints: I draw a ΔABQ with equal area of quadrilateral ABCD. Taking BQ as base of ΔABC I draw another triangle with same base and between two parallels of which one angle is 30°.
Solution:
Construction: From AY straight line a 5 cm long segment, AB is cut off. On A and B. points respectively perpendiculars AX and BM are drawn. From AX a 7 cm long segment AD and from BM a 4 cm long segment BC are cut. C and D are joined. Consequently, quadrilateral ABCD is formed.
Equal in area to □ABCD ΔADP is drawn. From point D parallel to AP a straight line DN is drawn. On point A, ∠PAF = 30° is drawn. AF cuts DN at point E.
Joining E, P, ΔAPE is drawn whose area is equal to □ABCD and one angle is 30°.
Proof: ΔAPD = ΔBCD
Again, ΔAPD and ΔAPE lie on same base AP and in between parallel lines AP and DE.
∴ ΔAPD = ΔAPE
∴ ΔAPE = □ABCD (Both are equal to ΔAPD)
Question 8. I draw any pentagon ABCDE and draw a triangle equal in area to it, of which one vertex is C.
Solution:
ABCDE, a pentagon is drawn. AC and CE diagonals are drawn. B From the point parallel to CA a straight line is drawn which cuts extended EA at point P.
Again, from point D parallel to CE a straight line is drawn which cuts extended AE at point Q C, P and C, Q are joined. ΔCPQ is the required triangle equal in area to pentagon ABCDE.
