Question

. AB= 8.5 cm, BC = 4.5 cm, CD=5 cm, BD= 7.5 cm, AD =6 cm ആയ ചതുർഭുജം
ABCD നിർമ്മിക്കുക. ABCD ക്ക് തുല്യപരപ്പളവുള്ള ത്രികാണം വരക്കുക

Answer 1

Answer:

I'm not understanding your language please write in english!!!!

Answer 2

A quadrilateral is a polygon with four sides.

Then let's plot it, check it below.

1) Since 5 line segments were given for a quadrilateral, one of them is an interior one. In this quadrilateral, a rhombus. We have a diagonal, id est a line segment between non-consecutive points.

2) Let's calculate the area of this rhombus. Since this polygon is made up of two triangles let's find it using Heron's Formula, not very popular. But equally valid, also we don't have the height nor angles.

All we need is the semi-perimeter, (half of the Perimeter (2P) and plug it in the formula:

\begin{gathered}\bigtriangleup ABD semi-perimeter:\frac{6+8.5+7.5}{2} \therefore s=11\\Area \:\bigtriangleup ABD=\sqrt{11(11-6)(11-8.5)(11-7.5)}=\frac{5\sqrt{77}}{2} \approx21.94 cm\\\\\bigtriangleup \:BCD\: semi-perimeter:\frac{4.5+5+7.5}{2} \therefore s=8.5\\\\Area \bigtriangleup BCD=\sqrt{8.5(8.5-4.5)(8.5-5)(8.5-7.5)}=\sqrt{119}\approx 10.90\end{gathered}

△ABDsemi−perimeter:

2

6+8.5+7.5

∴s=11

Area△ABD=

11(11−6)(11−8.5)(11−7.5)

=

2

5

77

≈21.94cm

△BCDsemi−perimeter:

2

4.5+5+7.5

∴s=8.5

Area△BCD=

8.5(8.5−4.5)(8.5−5)(8.5−7.5)

=

119

≈10.90

Area\: of\: Rhombus\: ABCD=21.94+10.90=32.84 cm^2AreaofRhombusABCD=21.94+10.90=32.84cm

2

3) Well, now we need to trace a triangle whose area is 32.84 cm^2. From the classical formula for Area of Triangles we can write:

\frac{b*h}{2}=32.84 \therefore b*h=65.68

2

b∗h

=32.84∴b∗h=65.68

Let's find out two values one for the base and another for height. Since 65.58 can be divided both by two and three, it is divisible by 6.

So

\begin{gathered}65.58 : 6 =10.93\\b=6\\h=10.93\end{gathered}

65.58:6=10.93

b=6

h=10.93

ഇത്‌ correct ആണോന്ന് അറിയില്ലാട്ടോ ☹️

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