Class 6 - Mathematics
Chapter - Practical Geometry : Exercise 10.6
Top Block 1
Below are given the measures of certain sides and angles of triangles. Identify those which cannot be constructed and, say why you cannot construct them. Construct rest of the triangle.
Triangle Given measurements
Question : 1. ΔABC𝓂∠A = 85o ; 𝓂∠B = 115o ; AB = 5 cm
Answer :
In ΔABC, 𝓂∠A = 85o,𝓂∠B = 115o, AB = 5 cm Construction of ΔABC is not possible because 𝓂∠A = 85o+𝓂∠B = 200o, and we know that the sum of angles of a triangle should be 180o.
Question : 2. ΔPQR 𝓂∠Q = 30o ; 𝓂∠R = 60o ; QR = 4.7 cm
Answer :
To construct: ΔPQR where 𝓂∠Q = 30o,𝓂∠R = 60o and QR = 4.7 cm.
Draw a line segment QR = 4.7 cm.
(a) At point Q, draw ∠XQR = 30o with the help of compass.
(b) At point R, draw ∠YRQ = 60o with the help of compass.
(c) QX and RY intersect at point P.
It is the required triangle PQR.
Question : 3. ΔABC𝓂∠A = 70o ; 𝓂∠B = 50o ; AC = 3 cm
Answer :
We know that the sum of angles of a triangle is 180o.
∴ 𝓂∠A + 𝓂∠B + 𝓂∠C = 180o
⇒ 70o+50o+𝓂∠C = 180o
⇒ 120o + 𝓂∠C = 180o
⇒𝓂∠C = 180o– 120o
⇒𝓂∠C = 60o
To construct: ΔABC where 𝓂∠A = 70o, 𝓂∠C = 60o and AC = 3 cm.
Mddle block 1
(a) Draw a line segment AC = 3 cm.
(b) At point C, draw ∠YCA = 60o.
(c) At point A, draw ∠XAC = 70o.
(d) Rays XA and YC intersect at point B
It is the required triangle ABC.
Question : 4. ΔLMN𝓂∠L = 60o ; 𝓂∠N = 120o ; LM = 5 cm
Answer :
In ΔLMN , 𝓂∠L = 60o,𝓂∠N = 120o, LM = 5 cm
This ΔLMN is not possible to construct because 𝓂∠L + 𝓂∠N = 60o+120o=180o which forms a linear pair.
Question : 5. ΔABC BC = 2 cm; AB = 4 cm; AC = 2 cm
Answer :
ΔABC, BC = 2 cm, AB = 4 cm and AC = 2 cm
This ΔABC is not possible to construct because the condition is
Sum of lengths of two sides of a triangle should be greater than the third side.
Question : 6. ΔPQR PQ = 3.5 cm; QR = 4 cm; PR = 3.5 cm
Answer :
To construct: ΔPQR where PQ = 3.5 cm, QR = 4 cm and PR = 3.5 cm
Steps of construction:
(a) Draw a line segment QR = 4 cm.
(b) Taking Q as centre and radius 3.5 cm, draw an arc.
(c) Similarly, taking R as centre and radius 3.5 cm, draw an another arc which intersects the first arc at point P.
It is the required triangle PQR.
Question : 7. ΔXYZ XY = 3 cm; YZ = 4 cm; XZ = 5 cm
Answer :
To construct: A triangle whose sides are XY = 3 cm, YZ = 4 cm and XZ = 5 cm.
(a) Draw a line segment ZY = 4 cm.
(b) Taking Z as centre and radius 5 cm, draw an arc.
(c) Taking Y as centre and radius 3 cm, draw another arc.
(d) Both arcs intersect at point X.
It is the required triangle XYZ.
Question : 8. ΔDEF DE = 4.5 cm; EF = 5.5 cm; DF = 4 cm
Answer :
To construct: A triangle DEF whose sides are DE = 4.5 cm, EF = 5.5 cm and DF = 4 cm.
(a) Draw a line segment EF = 5.5 cm.
(b) Taking E as centre and radius 4.5 cm, draw an arc.
(a) Taking F as centre and radius 4 cm, draw another arc which intersects the first arc at point D.
It is the required triangle DEF.