Problem Set 4 Geometry 9th Std Maths Part 2 Answers Chapter 4 Constructions of Triangles
Steps of construction:
i. Draw seg YZ of length 4.9 cm.
ii. Draw ray YT, such that ∠ZYT = 75°.
iii. Mark point W on ray YT such that l(YW) = 10.3 cm.
iv. Join points W and Z.
v. Draw perpendicular bisector of seg WZ intersecting ray YT. Name the point as X.
vi. Join the points X and Z.
Hence, ∆XYZ is the required triangle.
ii. In ∆ADB
AB = BD [From (i)]
∴ ∠BAD = ∠BDA = x° ….(iv) [Isosceles triangle theorem]
In ∆ABD, ∠ABC is the exterior angle.
∴ ∠BAD + ∠BDA = ∠ABC [Remote interior angles theorem]
x + x = 70° [From (iv)]
∴ 2x = 70° x = 35°
∴ ∠ADB = 35°
∴ ∠D = 35°
Similarly, ∠E = 30°
iii. Now, in ∆ADE
∠D = 35°, ∠E = 30° and DE = 11.2 cm
Elence, ∆ADE can be drawn.
iv. Since, AB = BD
∴ Point B lies on perpendicular bisector of seg AD.
Also AC = CE
∴ Point C lies on perpendicular bisector of seg AE.
∴ Points B and C can be located by drawing the perpendicular bisector of AD and AE respectively.
∴ ∆ABC can be drawn.
Steps of construction:
i. Draw seg DE of length 11.2 cm.
ii. From point D draw ray making angle of 35°.
iii. From point E draw ray making angle of 30°.
iv. Name the point of intersection of two rays as A.
v. Draw the perpendicular bisector of seg DA and seg EA intersecting seg DE in B and C respectively.
vi. Join AB and AC.
Hence, ∆ABC is the required triangle.
Steps of construction:
i. Draw seg QR of length 6.4 cm.
ii. Draw ray QT, such that ∠RQT = 55°.
iii. Take point S on ray QT such that l(QS) = 2.4 cm.
iv. Join the points S and R.
v. Draw perpendicular bisector of seg SR intersecting ray QT.
Name that point as P.
vi. Join the points P and R.
Hence, ∆PQR is the required triangle.