(a) Construct a triangle ABC such that BC = 8 cm, AC = 10 cm and ∠ABC = 90°.
(b) Construct an incircle to this triangle. Mark the centre as I.
(c) Measure and write the length of the in-radius.
(d) Measure and write the length of the tangents from vertex C to the incircle.
(e) Mark points P, Q and R where the incircle touches the sides AB, BC, and AC of the triangle respectively. Write the relationship between ∠RIQ and ∠QCR.
(Use a ruler and a compass for this question.)

Question

(a) Construct a triangle ABC such that BC = 8 cm, AC = 10 cm and ∠ABC = 90°.
(b) Construct an incircle to this triangle. Mark the centre as I.
(c) Measure and write the length of the in-radius.
(d) Measure and write the length of the tangents from vertex C to the incircle.
(e) Mark points P, Q and R where the incircle touches the sides AB, BC, and AC of the triangle respectively. Write the relationship between ∠RIQ and ∠QCR.
(Use a ruler and a compass for this question.)

KnowledgeBoat · Accepted Answer

Steps of construction : 1. Draw a line segment BC = 8 cm. 2. Draw BX perpendicular to BC. 3. With C as center and radius = 10 cm, draw an arc cutting BX at A. 4. Join AB and AC. 5. Draw AW, BY and CZ the angle bisectors of A, B and C respectively. 6. Mark the point of intersection as I. 7. Draw IR perpendicular to side AC. 8. With I as center and radius IR draw a circle, which is the required incircle. 9. Mark points P, Q and R where the incircle touches the sides AB, BC, and AC of the triangle respectively. 10. Measure CQ and CR. From figure, ⇒ ∠IRC = ∠IQC = 90° (The radius from the center of the circle to the point of tangency is perpendicular to the tangent line.) ⇒ ∠RCI = ∠QCI = (As CZ is angle bisector) In △ IRC, ⇒ ∠RIC = 180° - ∠RCI - ∠IRC [∵ Sum of ∠'s in a Δ = 180°] ⇒ ∠RIC = 180° - - 90° ⇒ ∠RIC = 90° - ............(1) In △ IQC, ⇒ ∠QIC = 180° - ∠IQC - ∠ICQ [∵ Sum of ∠'s in a Δ = 180°] ⇒ ∠QIC = 180° - 90° - ⇒ ∠QIC = 90° - ............(2) Adding equations (1) and (2), we get : ⇒ ∠RIC + ∠QIC = 90° - + 90° - ⇒ ∠RIQ = 180° - ∠C ⇒ ∠RIQ = 180° - ∠RCQ ⇒ ∠RIQ + ∠RCQ = 180°. Hence, ∠RIQ + ∠QCR = 180°.

Class marks (Wages)	No. of workers
425	6
475	12
525	15
575	17
625	7
675	13

Mathematics

Constructions

Related Questions

In the given figure, angle ABC = 70° and angle ACB = 50°. Given, O is the centre of the circle and PT is the tangent to the circle. Then calculate the following angles
(a) ∠CBT
(b) ∠BAT
(c) ∠PBT
(d) ∠APT

The daily wages of workers in a construction unit were recorded as follows :
Class marks (Wages) No. of workers
425 6
475 12
525 15
575 17
625 7
675 13
Form a frequency distribution table with class intervals and find modal wage by plotting a histogram.

Mathematics

Constructions

Related Questions

In the given figure, angle ABC = 70° and angle ACB = 50°. Given, O is the centre of the circle and PT is the tangent to the circle. Then calculate the following angles(a) ∠CBT(b) ∠BAT(c) ∠PBT(d) ∠APT

The daily wages of workers in a construction unit were recorded as follows :Class marks (Wages)No. of workers4256475125251557517625767513Form a frequency distribution table with class intervals and find modal wage by plotting a histogram.

In the given figure, angle ABC = 70° and angle ACB = 50°. Given, O is the centre of the circle and PT is the tangent to the circle. Then calculate the following angles
(a) ∠CBT
(b) ∠BAT
(c) ∠PBT
(d) ∠APT

The daily wages of workers in a construction unit were recorded as follows :
Class marks (Wages) No. of workers
425 6
475 12
525 15
575 17
625 7
675 13
Form a frequency distribution table with class intervals and find modal wage by plotting a histogram.