Answer:
ASA postulate is used the triangles are congruent
Step-by-step explanation:
if we look carefully we find that the angle and side marked are equal . Also opposite angles are equal. so the lines bisecting are making opposite equal angles.
hence two angles and one side are equal and ASA (angle side angle) postulate is used. the triangles are congruent
Area of isosceles triangle is 63.98 square units.
<u>Solution:</u>
Note: Refer the image attached below
Given that leg of an isosceles triangle is 16; Measured of one of the angle is 150; Need to determine area of triangle.
Consider the figure ABC is required isosceles triangle
From given information,
AC = CB = 16 and ![\angle \mathrm{ACB}=150^{\circ}](https://tex.z-dn.net/?f=%5Cangle%20%5Cmathrm%7BACB%7D%3D150%5E%7B%5Ccirc%7D)
So we have two sides and angle between them. Let us use law of cosine
On applying law of cosine on triangle ABC we get,
![\begin{array}{l}{\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{CB}^{2}-2 \times \mathrm{AC} \times \mathrm{CB} \cos \mathrm{C}} \\\\ {=>\mathrm{AB}^{2}=16^{2}+16^{2}-2 \times 16 \times 16 \cos 150} \\\\ {\Rightarrow \mathrm{AB}^{2}=16^{2}+16^{2}+443.4} \\\\ {\Rightarrow \mathrm{AB}^{2}=955.4} \\\\ {\Rightarrow \mathrm{AB}=30.91}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%5Cmathrm%7BAB%7D%5E%7B2%7D%3D%5Cmathrm%7BAC%7D%5E%7B2%7D%2B%5Cmathrm%7BCB%7D%5E%7B2%7D-2%20%5Ctimes%20%5Cmathrm%7BAC%7D%20%5Ctimes%20%5Cmathrm%7BCB%7D%20%5Ccos%20%5Cmathrm%7BC%7D%7D%20%5C%5C%5C%5C%20%7B%3D%3E%5Cmathrm%7BAB%7D%5E%7B2%7D%3D16%5E%7B2%7D%2B16%5E%7B2%7D-2%20%5Ctimes%2016%20%5Ctimes%2016%20%5Ccos%20150%7D%20%5C%5C%5C%5C%20%7B%5CRightarrow%20%5Cmathrm%7BAB%7D%5E%7B2%7D%3D16%5E%7B2%7D%2B16%5E%7B2%7D%2B443.4%7D%20%5C%5C%5C%5C%20%7B%5CRightarrow%20%5Cmathrm%7BAB%7D%5E%7B2%7D%3D955.4%7D%20%5C%5C%5C%5C%20%7B%5CRightarrow%20%5Cmathrm%7BAB%7D%3D30.91%7D%5Cend%7Barray%7D)
Now we are having three sides of triangle,
AB = 30.91; BC = 16 and CA = 16
As three sides are given, we can apply heron’s formula to determine area of triangle ABC. According to herons formula,
![\begin{array}{l}{\text { Area of Triangle }=\sqrt{s(s-a)(s-b)(s-c)}} \\\\ {\text { Where } s=\frac{a+b+c}{2}}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%5Ctext%20%7B%20Area%20of%20Triangle%20%7D%3D%5Csqrt%7Bs%28s-a%29%28s-b%29%28s-c%29%7D%7D%20%5C%5C%5C%5C%20%7B%5Ctext%20%7B%20Where%20%7D%20s%3D%5Cfrac%7Ba%2Bb%2Bc%7D%7B2%7D%7D%5Cend%7Barray%7D)
In our case a = AB = 30.91; b = BC = 16; c = CA = 16
![\begin{array}{l}{\text { So } s=\frac{A B+B C+C A}{2}=\frac{30.91+16+16}{2}=31.455} \\\\ {\text { Area of Triangle } A B C=\sqrt{31.455(31.455-30.91)(31.455-16)(31.455-16)}} \\\\ {\Rightarrow \text { Area of Triangle } A B C=\sqrt{31.455 \times 0.545 \times 15.455 \times 15.455}=\sqrt{4094.72}=63.98}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%5Ctext%20%7B%20So%20%7D%20s%3D%5Cfrac%7BA%20B%2BB%20C%2BC%20A%7D%7B2%7D%3D%5Cfrac%7B30.91%2B16%2B16%7D%7B2%7D%3D31.455%7D%20%5C%5C%5C%5C%20%7B%5Ctext%20%7B%20Area%20of%20Triangle%20%7D%20A%20B%20C%3D%5Csqrt%7B31.455%2831.455-30.91%29%2831.455-16%29%2831.455-16%29%7D%7D%20%5C%5C%5C%5C%20%7B%5CRightarrow%20%5Ctext%20%7B%20Area%20of%20Triangle%20%7D%20A%20B%20C%3D%5Csqrt%7B31.455%20%5Ctimes%200.545%20%5Ctimes%2015.455%20%5Ctimes%2015.455%7D%3D%5Csqrt%7B4094.72%7D%3D63.98%7D%5Cend%7Barray%7D)
Answer:
Step-by-step explanation:
The question looks like this.
square root (3111)^2
The square root of 3111 is 55.78
Now you have to square it 55.78^2 = 3111.4
The 0.4 comes up because 55.78 is a round of the actual square root which is
55.776339069537362385549858627874
This when squared gives you 3111 to the limit of the calculator used.
Answer:
Step-by-step explanation:
a+c=300
3a+c=2100