Find The area of triangle base-5cm, height-4cm
Answers
Answer:
area=10cm²
Step-by-step explanation:
area of ∆ = 1/2*b*h
=1/2*5*4
=10cm²
Step-by-step explanation:
There are two ways of drawing a triangle similar to isosceles triangle, whose base is 5 cm and height 4 cm.
First you can find the other two sides using Pythagoras theorem and then find \frac{2}{3}
3
2
of each side and construct the triangle.
Let other two equal sides of triangle be x cm.
As, Length of perpendicular on side BC=AM=4 cm
In an Isosceles triangle, perpendicular acts as perpendicular bisector.
So, BM=MC=2.5 cm
So, In Δ AMB, using Pythagoras theorem
AB²=AM²+MB²
AB²=4²+(2.5)²
AB²=16+4.25
=20.25
AB=4.5 cm
So, AB=AC=4.5 cm
\frac{2\times AB}{3}=\frac{2\times AC}{3}=\frac{2\times 4.5}{3}=3 cm
3
2×AB
=
3
2×AC
=
3
2×4.5
=3cm
\frac{2\times CB}{3}=\frac{10}{3}
3
2×CB
=
3
10
So, If ΔPQR~ΔABC by SSS, then
Following procedure is adopted.
Draw line segment BC of length 5 cm.
From B and C mark arc at a distance of 4.5 on one side of line segment BC.The point of intersection of arcs is point A.
Now, to draw ΔPQR similar to ΔABC,
draw a line segment QR equal to \frac{2\times CB}{3}=\frac{10}{3}
3
2×CB
=
3
10
.
From both ends of segment QR, that is Q and R,Such that PQ=QR, mark arc at a distance of
\frac{2\times AB}{3}=\frac{2\times AC}{3}=\frac{2\times 4.5}{3}=3 cm
3
2×AB
=
3
2×AC
=
3
2×4.5
=3cm on one side of line segment QR, such that their point of intersection is point P.
This is required triangle PQR, similar to Triangle ABC.
Second Method
Draw line segment BC of length 5 cm.
Draw perpendicular bisector of side BC, by opening the compass more than half of length of side BC, the point of intersection of these arcs on any side is point A.
This is required Δ ABC.
Now draw ray from point B, such that ∠CBP=acute angle.
Mark 5 arcs at equal distances from point B.The last arc is at Point P.
Join CP.
At point P draw semicircular arc, you can draw a circle also.
Measure the length of arc TS.
At point Q with the same measurement,as at point P, draw same semicircular arc, and from point O, cut arc equal to arc TS which is OV.
Draw segment from Q passing through V, cutting the segment BC at R.
This is required ΔBQR similar to ΔABC.
