Construct an isosceles triangle whose base is 7 cm and altitude 6 cm and then another triangle whose sides are one and half times of the given triangle.
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Solution:
Draw the line segment of the base 8 cm. Draw perpendicular bisector of the line. Mark a point on the bisector which measures 4 cm from the base. Connect this point from both ends.
Then draw another line that makes an acute angle with the given line. Divide the line into m + n parts where m and n are the ratios given.
Two triangles are said to be similar if their corresponding angles are equal. They are said to satisfy Angle-Angle-Angle (AAA) Axiom.
The basic proportionality theorem states that “If a straight line is drawn parallel to a side of a triangle, then it divides the other two sides proportionally".
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are times the corresponding sides of the isosceles triangle.
Steps of construction:
Draw BC = 8cm.
Through D, the mid-point of BC, draw the perpendicular to BC and cut an arc from D on it such that DA = 4cm. Join BA and CA. ΔABC is obtained.
Draw the ray BX so that ∠CBX is acute.
Mark 3 (since, 3 > 2 in 1
1
2
= 3/2) points B₁, B₂, B₃ on BX such that BB₁ = B₁B₂ = B₂B₃
Join B₂ (2nd point ∵ 2 < 3) to C and draw B₃C' parallel to B₂C, intersecting BC extended at C’.
Through C’ draw C'A' parallel to CA to intersect BA extended to A’. Now, ΔA'BC' is the required triangle similar to ΔABC where BA'/BA = C'A'/CA = BC'/BC = 3/2
Proof:
In ΔBB₃C', B₂C || B₃C',
Hence by Basic proportionality theorem,
B₂B₃/BB₂ = CC'/BC = 1/2
Adding 1 to CC'/BC = 1/2
CC'/BC + 1 = 1/2 + 1
(BC+CC')/BC = 3/2
BC'/BC = 3/2
Consider ΔBAC and ΔBA'C'
∠ABC = ∠A'BC' (Common)
∠BCA = ∠BC'A' (Corresponding angles ∵ CA || C'A')
∠BAC = ∠BA'C' (Corresponding angles)
By AAA axiom, ΔBAC ~ ΔBA'C'
∴ Corresponding sides are proportional
Hence,
BA'/BA = BC'/BC= C'A'/CA = 3/2
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Answer:
Draw a base BC of length 8 cm.
2. Draw the perpendicular bisector of BC and name it PQ. Let it intersect BC in M.
3. From M, mark an arc A at a distance of 4 cm on PQ.
4. Join B−A and C−B. △ABC is the first isosceles triangle.
5. Extend BC and BA. Place the pointed end of the compass at C and mark an arc of radius CM on the opposite side of M on the extended segment BC. Mark the point as D (Since CM is half of BC, BD will be BC+0.5BC=1.5BC)
6. From D, draw a line parallel to AC and let it intersect the extended segment AB at E.
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